Proving a continuous function is a constant











up vote
1
down vote

favorite












Let $f$ be a function that is continuous on an interval $[a,b]$ and differentiable on $(a,b)$. Assume $f'(x)=0$ for all $xin (a,b)$. Show f is a constant.



I feel like the obvious set up is by contradiction. If I assume $f$ is not a constant, then it must have a variable, but I'm unsure how to construct an $f$ (without loss of generality) which I can then differentiate to contradict $f'(x)=0$.



any suggestions to start this proof?










share|cite|improve this question




















  • 3




    Do you know the mean value theorem?
    – Randall
    Nov 29 at 4:07










  • doesn't that rely on the assumption that $f(a)=f(b)$ though?
    – Jess Savoie
    Nov 29 at 4:10










  • No. You may be thinking of Rolle.
    – Randall
    Nov 29 at 4:11










  • Why don't we try using the very definition of differentiation?
    – Aniruddha Deshmukh
    Nov 29 at 4:11






  • 2




    @AniruddhaDeshmukh: Because that would be much more difficult than using the mean value theorem. See for example, Tim Gowers's discussion here: dpmms.cam.ac.uk/~wtg10/meanvalue.html
    – Hans Lundmark
    Nov 29 at 5:50















up vote
1
down vote

favorite












Let $f$ be a function that is continuous on an interval $[a,b]$ and differentiable on $(a,b)$. Assume $f'(x)=0$ for all $xin (a,b)$. Show f is a constant.



I feel like the obvious set up is by contradiction. If I assume $f$ is not a constant, then it must have a variable, but I'm unsure how to construct an $f$ (without loss of generality) which I can then differentiate to contradict $f'(x)=0$.



any suggestions to start this proof?










share|cite|improve this question




















  • 3




    Do you know the mean value theorem?
    – Randall
    Nov 29 at 4:07










  • doesn't that rely on the assumption that $f(a)=f(b)$ though?
    – Jess Savoie
    Nov 29 at 4:10










  • No. You may be thinking of Rolle.
    – Randall
    Nov 29 at 4:11










  • Why don't we try using the very definition of differentiation?
    – Aniruddha Deshmukh
    Nov 29 at 4:11






  • 2




    @AniruddhaDeshmukh: Because that would be much more difficult than using the mean value theorem. See for example, Tim Gowers's discussion here: dpmms.cam.ac.uk/~wtg10/meanvalue.html
    – Hans Lundmark
    Nov 29 at 5:50













up vote
1
down vote

favorite









up vote
1
down vote

favorite











Let $f$ be a function that is continuous on an interval $[a,b]$ and differentiable on $(a,b)$. Assume $f'(x)=0$ for all $xin (a,b)$. Show f is a constant.



I feel like the obvious set up is by contradiction. If I assume $f$ is not a constant, then it must have a variable, but I'm unsure how to construct an $f$ (without loss of generality) which I can then differentiate to contradict $f'(x)=0$.



any suggestions to start this proof?










share|cite|improve this question















Let $f$ be a function that is continuous on an interval $[a,b]$ and differentiable on $(a,b)$. Assume $f'(x)=0$ for all $xin (a,b)$. Show f is a constant.



I feel like the obvious set up is by contradiction. If I assume $f$ is not a constant, then it must have a variable, but I'm unsure how to construct an $f$ (without loss of generality) which I can then differentiate to contradict $f'(x)=0$.



any suggestions to start this proof?







calculus real-analysis continuity proof-writing






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 at 4:07

























asked Nov 29 at 4:06









Jess Savoie

487




487








  • 3




    Do you know the mean value theorem?
    – Randall
    Nov 29 at 4:07










  • doesn't that rely on the assumption that $f(a)=f(b)$ though?
    – Jess Savoie
    Nov 29 at 4:10










  • No. You may be thinking of Rolle.
    – Randall
    Nov 29 at 4:11










  • Why don't we try using the very definition of differentiation?
    – Aniruddha Deshmukh
    Nov 29 at 4:11






  • 2




    @AniruddhaDeshmukh: Because that would be much more difficult than using the mean value theorem. See for example, Tim Gowers's discussion here: dpmms.cam.ac.uk/~wtg10/meanvalue.html
    – Hans Lundmark
    Nov 29 at 5:50














  • 3




    Do you know the mean value theorem?
    – Randall
    Nov 29 at 4:07










  • doesn't that rely on the assumption that $f(a)=f(b)$ though?
    – Jess Savoie
    Nov 29 at 4:10










  • No. You may be thinking of Rolle.
    – Randall
    Nov 29 at 4:11










  • Why don't we try using the very definition of differentiation?
    – Aniruddha Deshmukh
    Nov 29 at 4:11






  • 2




    @AniruddhaDeshmukh: Because that would be much more difficult than using the mean value theorem. See for example, Tim Gowers's discussion here: dpmms.cam.ac.uk/~wtg10/meanvalue.html
    – Hans Lundmark
    Nov 29 at 5:50








3




3




Do you know the mean value theorem?
– Randall
Nov 29 at 4:07




Do you know the mean value theorem?
– Randall
Nov 29 at 4:07












doesn't that rely on the assumption that $f(a)=f(b)$ though?
– Jess Savoie
Nov 29 at 4:10




doesn't that rely on the assumption that $f(a)=f(b)$ though?
– Jess Savoie
Nov 29 at 4:10












No. You may be thinking of Rolle.
– Randall
Nov 29 at 4:11




No. You may be thinking of Rolle.
– Randall
Nov 29 at 4:11












Why don't we try using the very definition of differentiation?
– Aniruddha Deshmukh
Nov 29 at 4:11




Why don't we try using the very definition of differentiation?
– Aniruddha Deshmukh
Nov 29 at 4:11




2




2




@AniruddhaDeshmukh: Because that would be much more difficult than using the mean value theorem. See for example, Tim Gowers's discussion here: dpmms.cam.ac.uk/~wtg10/meanvalue.html
– Hans Lundmark
Nov 29 at 5:50




@AniruddhaDeshmukh: Because that would be much more difficult than using the mean value theorem. See for example, Tim Gowers's discussion here: dpmms.cam.ac.uk/~wtg10/meanvalue.html
– Hans Lundmark
Nov 29 at 5:50










5 Answers
5






active

oldest

votes

















up vote
3
down vote













Let, $x,yin [a,b]$ then by Mean value theorem we have,
$$f(x)-f(y)=(x-y)f'(xi)$$ for some $x<xi<y$ and as given $f'(t)=0,forall tin(a,b)$ so, $f'(xi)=0implies f(x)=f(y),forall x,yin[a,b]$



Hence, $f$- is constant.






share|cite|improve this answer




























    up vote
    0
    down vote













    A proof by contradiction:



    We have to show: $$(f'(x)=0 implies f(x)=C) ni text{C is a constant}$$



    Which is nothing but, $$Pimplies Q$$.



    Suppose: $$lnot (Pimplies Q)$$



    Which is nothing but, $$P land lnot Q$$



    Therefore, suppose:
    $$f'(x)=0 land f(x)=h(x) ni text{h(x) is an arbitrary non-constant function}$$



    [$(h'(x)neq 0)(forall xneq C^p_n)$; where $C^p$ is the critical point, and $n$ is an arbitrary index.] This bit is not really relevant since, we are comparing $f(x)$ and $h(x)$ on $Asubseteq D$ where the functions are identical: f(x)=h(x).



    Notice, however, $$frac{d}{dx}f(x)=frac{d}{dx}h(x)$$



    $$frac{df}{dx}=frac{dh}{dx}$$



    Recall that we assumed: $h(x) text{ is an arbitrary non-constant function}$



    Therefore,



    $$frac{dh}{dx}=[k(x) lor L]$$
    Where $k(x)$ is an arbitrary function not equal to zero and $L$ is an arbitrary non-zero constant.



    However, in either case:



    $$frac{dh}{dx}neq0=frac{df}{dx}=f'(x)$$



    Therefore, since $lnot (Pimplies Q) implies bot$, then:



    $$(Pimplies Q)implies top$$






    share|cite|improve this answer



















    • 1




      I'm not sure all the logic actually helps. Also, the non-constant function $h(x) = begin{cases} 1, & x geq 0\ -1, & x<0end{cases}$ has $h'(x)=0$ (at all but one point) but isn't constant, which complicates your argument.
      – Randall
      Nov 29 at 15:02












    • @Randall I thought the logical mechanism might be useful. That said, $(h'(x)neq 0)(forall xneq C^p_n)$; where $C_p$ is the critical point, and $n$ is an arbitrary index. Other than that, thanks for the input.
      – Bertrand Wittgenstein's Ghost
      Nov 29 at 18:16


















    up vote
    0
    down vote














    Let f be a function that is continuous on an interval [a,b] and differentiable on (a,b)$




    1) that means for all $a le a' < b' le b$ then $f$ is continuous on the interval $[a', b']subset [a,b]$ and differentiable on $(a',b') subset (a,b)$.



    and



    2) The mean value thereom applies.



    So for any $a le a' < b' le b$ we have that there is a $c; a' < c < b'$ where $f'(c) = frac {f(b') - f(a')}2$.



    But $f'(c) = 0$ so $f(b') = f(a')$ for all $a le a' < b' le b$.



    So $f$ is constant in the interval $[a,b]$ (can't say anythin about $f$ outside that interval.)






    share|cite|improve this answer




























      up vote
      0
      down vote













      Here is an alternate proof which bypasses the Mean Value Theorem, and which I hope might be more conducive to visualization. First, I'll start off with a small topological lemma to abstract out that part of the argument:



      Lemma: Suppose $A subseteq [0, 1]$ satisfies $0 in A$, $A$ is closed under the left half-interval topology, and $A$ is open under the right half-interval topology. Then $A = [0, 1]$.



      Proof (outline): Let $B := { x in [0, 1] mid [0, x] subseteq A }$ and $c := sup B$ (using the fact that $0 in B$ so $B$ is nonempty) and suppose that $c < 1$. Then from the condition that $A$ is closed under the left half-interval topology, we can conclude that $c in B$. On the other hand, the fact that $A$ is open under the right half-interval topology will now give a contradiction. Therefore, $c = 1$; and again, from the condition that $A$ is closed under the left half-interval topology, this will imply that $1 in B$, so $A = [0, 1]$ as desired. $square$



      Now, to use this: to illustrate the essential points of the argument, I will restrict to the case where the domain of $f$ is $[0, 1]$ and $f(0) = 0$. The more general case should be straightforward to prove in a similar manner. Now, fix any $epsilon > 0$, and let $A_epsilon := { x in [0, 1] mid |f(x)| le epsilon x }$. Then $0 in A_epsilon$; and it should be straightforward to see that $A_epsilon$ is closed in the usual topology, and therefore also closed in the left half-interval topology. On the other hand, suppose $x in A_epsilon$; then from the hypothesis $f'(x) = 0$, we see that there exists $delta > 0$ such that whenever $0 < |y-x| < delta$, then $left| frac{f(y) - f(x)}{y - x} right| < epsilon$. Then, if in addition $y > x$, it follows that $|f(y)| le |f(x)| + |f(y) - f(x)| < epsilon x + epsilon |y-x| = epsilon y$; this shows that $A_epsilon$ is open in the right half-interval topology. Now from the lemma, it follows that $A_epsilon = [0, 1]$.



      In summary, we have shown that $forall epsilon > 0, forall x in [0, 1], |f(x)| le epsilon x$. We may now interchange the two quantifiers to conclude $forall x in [0, 1], forall epsilon > 0, |f(x)| le epsilon x$. However, this easily implies $forall x in [0, 1], f(x) = 0$.






      share|cite|improve this answer




























        up vote
        -1
        down vote













        Well, $f'(x)$ gives the rate of change of $f$ at a point $x$. If $f'(x)=0 forall x in (a,b)$, then $f(x)$ isn't changing, i.e. it is constant, on $(a,b)$.



        That's the idea, you just need to show it mathematically.






        share|cite|improve this answer





















          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3018160%2fproving-a-continuous-function-is-a-constant%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          5 Answers
          5






          active

          oldest

          votes








          5 Answers
          5






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          3
          down vote













          Let, $x,yin [a,b]$ then by Mean value theorem we have,
          $$f(x)-f(y)=(x-y)f'(xi)$$ for some $x<xi<y$ and as given $f'(t)=0,forall tin(a,b)$ so, $f'(xi)=0implies f(x)=f(y),forall x,yin[a,b]$



          Hence, $f$- is constant.






          share|cite|improve this answer

























            up vote
            3
            down vote













            Let, $x,yin [a,b]$ then by Mean value theorem we have,
            $$f(x)-f(y)=(x-y)f'(xi)$$ for some $x<xi<y$ and as given $f'(t)=0,forall tin(a,b)$ so, $f'(xi)=0implies f(x)=f(y),forall x,yin[a,b]$



            Hence, $f$- is constant.






            share|cite|improve this answer























              up vote
              3
              down vote










              up vote
              3
              down vote









              Let, $x,yin [a,b]$ then by Mean value theorem we have,
              $$f(x)-f(y)=(x-y)f'(xi)$$ for some $x<xi<y$ and as given $f'(t)=0,forall tin(a,b)$ so, $f'(xi)=0implies f(x)=f(y),forall x,yin[a,b]$



              Hence, $f$- is constant.






              share|cite|improve this answer












              Let, $x,yin [a,b]$ then by Mean value theorem we have,
              $$f(x)-f(y)=(x-y)f'(xi)$$ for some $x<xi<y$ and as given $f'(t)=0,forall tin(a,b)$ so, $f'(xi)=0implies f(x)=f(y),forall x,yin[a,b]$



              Hence, $f$- is constant.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 29 at 4:11









              Sujit Bhattacharyya

              945317




              945317






















                  up vote
                  0
                  down vote













                  A proof by contradiction:



                  We have to show: $$(f'(x)=0 implies f(x)=C) ni text{C is a constant}$$



                  Which is nothing but, $$Pimplies Q$$.



                  Suppose: $$lnot (Pimplies Q)$$



                  Which is nothing but, $$P land lnot Q$$



                  Therefore, suppose:
                  $$f'(x)=0 land f(x)=h(x) ni text{h(x) is an arbitrary non-constant function}$$



                  [$(h'(x)neq 0)(forall xneq C^p_n)$; where $C^p$ is the critical point, and $n$ is an arbitrary index.] This bit is not really relevant since, we are comparing $f(x)$ and $h(x)$ on $Asubseteq D$ where the functions are identical: f(x)=h(x).



                  Notice, however, $$frac{d}{dx}f(x)=frac{d}{dx}h(x)$$



                  $$frac{df}{dx}=frac{dh}{dx}$$



                  Recall that we assumed: $h(x) text{ is an arbitrary non-constant function}$



                  Therefore,



                  $$frac{dh}{dx}=[k(x) lor L]$$
                  Where $k(x)$ is an arbitrary function not equal to zero and $L$ is an arbitrary non-zero constant.



                  However, in either case:



                  $$frac{dh}{dx}neq0=frac{df}{dx}=f'(x)$$



                  Therefore, since $lnot (Pimplies Q) implies bot$, then:



                  $$(Pimplies Q)implies top$$






                  share|cite|improve this answer



















                  • 1




                    I'm not sure all the logic actually helps. Also, the non-constant function $h(x) = begin{cases} 1, & x geq 0\ -1, & x<0end{cases}$ has $h'(x)=0$ (at all but one point) but isn't constant, which complicates your argument.
                    – Randall
                    Nov 29 at 15:02












                  • @Randall I thought the logical mechanism might be useful. That said, $(h'(x)neq 0)(forall xneq C^p_n)$; where $C_p$ is the critical point, and $n$ is an arbitrary index. Other than that, thanks for the input.
                    – Bertrand Wittgenstein's Ghost
                    Nov 29 at 18:16















                  up vote
                  0
                  down vote













                  A proof by contradiction:



                  We have to show: $$(f'(x)=0 implies f(x)=C) ni text{C is a constant}$$



                  Which is nothing but, $$Pimplies Q$$.



                  Suppose: $$lnot (Pimplies Q)$$



                  Which is nothing but, $$P land lnot Q$$



                  Therefore, suppose:
                  $$f'(x)=0 land f(x)=h(x) ni text{h(x) is an arbitrary non-constant function}$$



                  [$(h'(x)neq 0)(forall xneq C^p_n)$; where $C^p$ is the critical point, and $n$ is an arbitrary index.] This bit is not really relevant since, we are comparing $f(x)$ and $h(x)$ on $Asubseteq D$ where the functions are identical: f(x)=h(x).



                  Notice, however, $$frac{d}{dx}f(x)=frac{d}{dx}h(x)$$



                  $$frac{df}{dx}=frac{dh}{dx}$$



                  Recall that we assumed: $h(x) text{ is an arbitrary non-constant function}$



                  Therefore,



                  $$frac{dh}{dx}=[k(x) lor L]$$
                  Where $k(x)$ is an arbitrary function not equal to zero and $L$ is an arbitrary non-zero constant.



                  However, in either case:



                  $$frac{dh}{dx}neq0=frac{df}{dx}=f'(x)$$



                  Therefore, since $lnot (Pimplies Q) implies bot$, then:



                  $$(Pimplies Q)implies top$$






                  share|cite|improve this answer



















                  • 1




                    I'm not sure all the logic actually helps. Also, the non-constant function $h(x) = begin{cases} 1, & x geq 0\ -1, & x<0end{cases}$ has $h'(x)=0$ (at all but one point) but isn't constant, which complicates your argument.
                    – Randall
                    Nov 29 at 15:02












                  • @Randall I thought the logical mechanism might be useful. That said, $(h'(x)neq 0)(forall xneq C^p_n)$; where $C_p$ is the critical point, and $n$ is an arbitrary index. Other than that, thanks for the input.
                    – Bertrand Wittgenstein's Ghost
                    Nov 29 at 18:16













                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  A proof by contradiction:



                  We have to show: $$(f'(x)=0 implies f(x)=C) ni text{C is a constant}$$



                  Which is nothing but, $$Pimplies Q$$.



                  Suppose: $$lnot (Pimplies Q)$$



                  Which is nothing but, $$P land lnot Q$$



                  Therefore, suppose:
                  $$f'(x)=0 land f(x)=h(x) ni text{h(x) is an arbitrary non-constant function}$$



                  [$(h'(x)neq 0)(forall xneq C^p_n)$; where $C^p$ is the critical point, and $n$ is an arbitrary index.] This bit is not really relevant since, we are comparing $f(x)$ and $h(x)$ on $Asubseteq D$ where the functions are identical: f(x)=h(x).



                  Notice, however, $$frac{d}{dx}f(x)=frac{d}{dx}h(x)$$



                  $$frac{df}{dx}=frac{dh}{dx}$$



                  Recall that we assumed: $h(x) text{ is an arbitrary non-constant function}$



                  Therefore,



                  $$frac{dh}{dx}=[k(x) lor L]$$
                  Where $k(x)$ is an arbitrary function not equal to zero and $L$ is an arbitrary non-zero constant.



                  However, in either case:



                  $$frac{dh}{dx}neq0=frac{df}{dx}=f'(x)$$



                  Therefore, since $lnot (Pimplies Q) implies bot$, then:



                  $$(Pimplies Q)implies top$$






                  share|cite|improve this answer














                  A proof by contradiction:



                  We have to show: $$(f'(x)=0 implies f(x)=C) ni text{C is a constant}$$



                  Which is nothing but, $$Pimplies Q$$.



                  Suppose: $$lnot (Pimplies Q)$$



                  Which is nothing but, $$P land lnot Q$$



                  Therefore, suppose:
                  $$f'(x)=0 land f(x)=h(x) ni text{h(x) is an arbitrary non-constant function}$$



                  [$(h'(x)neq 0)(forall xneq C^p_n)$; where $C^p$ is the critical point, and $n$ is an arbitrary index.] This bit is not really relevant since, we are comparing $f(x)$ and $h(x)$ on $Asubseteq D$ where the functions are identical: f(x)=h(x).



                  Notice, however, $$frac{d}{dx}f(x)=frac{d}{dx}h(x)$$



                  $$frac{df}{dx}=frac{dh}{dx}$$



                  Recall that we assumed: $h(x) text{ is an arbitrary non-constant function}$



                  Therefore,



                  $$frac{dh}{dx}=[k(x) lor L]$$
                  Where $k(x)$ is an arbitrary function not equal to zero and $L$ is an arbitrary non-zero constant.



                  However, in either case:



                  $$frac{dh}{dx}neq0=frac{df}{dx}=f'(x)$$



                  Therefore, since $lnot (Pimplies Q) implies bot$, then:



                  $$(Pimplies Q)implies top$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 30 at 0:40

























                  answered Nov 29 at 7:06









                  Bertrand Wittgenstein's Ghost

                  352114




                  352114








                  • 1




                    I'm not sure all the logic actually helps. Also, the non-constant function $h(x) = begin{cases} 1, & x geq 0\ -1, & x<0end{cases}$ has $h'(x)=0$ (at all but one point) but isn't constant, which complicates your argument.
                    – Randall
                    Nov 29 at 15:02












                  • @Randall I thought the logical mechanism might be useful. That said, $(h'(x)neq 0)(forall xneq C^p_n)$; where $C_p$ is the critical point, and $n$ is an arbitrary index. Other than that, thanks for the input.
                    – Bertrand Wittgenstein's Ghost
                    Nov 29 at 18:16














                  • 1




                    I'm not sure all the logic actually helps. Also, the non-constant function $h(x) = begin{cases} 1, & x geq 0\ -1, & x<0end{cases}$ has $h'(x)=0$ (at all but one point) but isn't constant, which complicates your argument.
                    – Randall
                    Nov 29 at 15:02












                  • @Randall I thought the logical mechanism might be useful. That said, $(h'(x)neq 0)(forall xneq C^p_n)$; where $C_p$ is the critical point, and $n$ is an arbitrary index. Other than that, thanks for the input.
                    – Bertrand Wittgenstein's Ghost
                    Nov 29 at 18:16








                  1




                  1




                  I'm not sure all the logic actually helps. Also, the non-constant function $h(x) = begin{cases} 1, & x geq 0\ -1, & x<0end{cases}$ has $h'(x)=0$ (at all but one point) but isn't constant, which complicates your argument.
                  – Randall
                  Nov 29 at 15:02






                  I'm not sure all the logic actually helps. Also, the non-constant function $h(x) = begin{cases} 1, & x geq 0\ -1, & x<0end{cases}$ has $h'(x)=0$ (at all but one point) but isn't constant, which complicates your argument.
                  – Randall
                  Nov 29 at 15:02














                  @Randall I thought the logical mechanism might be useful. That said, $(h'(x)neq 0)(forall xneq C^p_n)$; where $C_p$ is the critical point, and $n$ is an arbitrary index. Other than that, thanks for the input.
                  – Bertrand Wittgenstein's Ghost
                  Nov 29 at 18:16




                  @Randall I thought the logical mechanism might be useful. That said, $(h'(x)neq 0)(forall xneq C^p_n)$; where $C_p$ is the critical point, and $n$ is an arbitrary index. Other than that, thanks for the input.
                  – Bertrand Wittgenstein's Ghost
                  Nov 29 at 18:16










                  up vote
                  0
                  down vote














                  Let f be a function that is continuous on an interval [a,b] and differentiable on (a,b)$




                  1) that means for all $a le a' < b' le b$ then $f$ is continuous on the interval $[a', b']subset [a,b]$ and differentiable on $(a',b') subset (a,b)$.



                  and



                  2) The mean value thereom applies.



                  So for any $a le a' < b' le b$ we have that there is a $c; a' < c < b'$ where $f'(c) = frac {f(b') - f(a')}2$.



                  But $f'(c) = 0$ so $f(b') = f(a')$ for all $a le a' < b' le b$.



                  So $f$ is constant in the interval $[a,b]$ (can't say anythin about $f$ outside that interval.)






                  share|cite|improve this answer

























                    up vote
                    0
                    down vote














                    Let f be a function that is continuous on an interval [a,b] and differentiable on (a,b)$




                    1) that means for all $a le a' < b' le b$ then $f$ is continuous on the interval $[a', b']subset [a,b]$ and differentiable on $(a',b') subset (a,b)$.



                    and



                    2) The mean value thereom applies.



                    So for any $a le a' < b' le b$ we have that there is a $c; a' < c < b'$ where $f'(c) = frac {f(b') - f(a')}2$.



                    But $f'(c) = 0$ so $f(b') = f(a')$ for all $a le a' < b' le b$.



                    So $f$ is constant in the interval $[a,b]$ (can't say anythin about $f$ outside that interval.)






                    share|cite|improve this answer























                      up vote
                      0
                      down vote










                      up vote
                      0
                      down vote










                      Let f be a function that is continuous on an interval [a,b] and differentiable on (a,b)$




                      1) that means for all $a le a' < b' le b$ then $f$ is continuous on the interval $[a', b']subset [a,b]$ and differentiable on $(a',b') subset (a,b)$.



                      and



                      2) The mean value thereom applies.



                      So for any $a le a' < b' le b$ we have that there is a $c; a' < c < b'$ where $f'(c) = frac {f(b') - f(a')}2$.



                      But $f'(c) = 0$ so $f(b') = f(a')$ for all $a le a' < b' le b$.



                      So $f$ is constant in the interval $[a,b]$ (can't say anythin about $f$ outside that interval.)






                      share|cite|improve this answer













                      Let f be a function that is continuous on an interval [a,b] and differentiable on (a,b)$




                      1) that means for all $a le a' < b' le b$ then $f$ is continuous on the interval $[a', b']subset [a,b]$ and differentiable on $(a',b') subset (a,b)$.



                      and



                      2) The mean value thereom applies.



                      So for any $a le a' < b' le b$ we have that there is a $c; a' < c < b'$ where $f'(c) = frac {f(b') - f(a')}2$.



                      But $f'(c) = 0$ so $f(b') = f(a')$ for all $a le a' < b' le b$.



                      So $f$ is constant in the interval $[a,b]$ (can't say anythin about $f$ outside that interval.)







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Nov 30 at 0:53









                      fleablood

                      68k22684




                      68k22684






















                          up vote
                          0
                          down vote













                          Here is an alternate proof which bypasses the Mean Value Theorem, and which I hope might be more conducive to visualization. First, I'll start off with a small topological lemma to abstract out that part of the argument:



                          Lemma: Suppose $A subseteq [0, 1]$ satisfies $0 in A$, $A$ is closed under the left half-interval topology, and $A$ is open under the right half-interval topology. Then $A = [0, 1]$.



                          Proof (outline): Let $B := { x in [0, 1] mid [0, x] subseteq A }$ and $c := sup B$ (using the fact that $0 in B$ so $B$ is nonempty) and suppose that $c < 1$. Then from the condition that $A$ is closed under the left half-interval topology, we can conclude that $c in B$. On the other hand, the fact that $A$ is open under the right half-interval topology will now give a contradiction. Therefore, $c = 1$; and again, from the condition that $A$ is closed under the left half-interval topology, this will imply that $1 in B$, so $A = [0, 1]$ as desired. $square$



                          Now, to use this: to illustrate the essential points of the argument, I will restrict to the case where the domain of $f$ is $[0, 1]$ and $f(0) = 0$. The more general case should be straightforward to prove in a similar manner. Now, fix any $epsilon > 0$, and let $A_epsilon := { x in [0, 1] mid |f(x)| le epsilon x }$. Then $0 in A_epsilon$; and it should be straightforward to see that $A_epsilon$ is closed in the usual topology, and therefore also closed in the left half-interval topology. On the other hand, suppose $x in A_epsilon$; then from the hypothesis $f'(x) = 0$, we see that there exists $delta > 0$ such that whenever $0 < |y-x| < delta$, then $left| frac{f(y) - f(x)}{y - x} right| < epsilon$. Then, if in addition $y > x$, it follows that $|f(y)| le |f(x)| + |f(y) - f(x)| < epsilon x + epsilon |y-x| = epsilon y$; this shows that $A_epsilon$ is open in the right half-interval topology. Now from the lemma, it follows that $A_epsilon = [0, 1]$.



                          In summary, we have shown that $forall epsilon > 0, forall x in [0, 1], |f(x)| le epsilon x$. We may now interchange the two quantifiers to conclude $forall x in [0, 1], forall epsilon > 0, |f(x)| le epsilon x$. However, this easily implies $forall x in [0, 1], f(x) = 0$.






                          share|cite|improve this answer

























                            up vote
                            0
                            down vote













                            Here is an alternate proof which bypasses the Mean Value Theorem, and which I hope might be more conducive to visualization. First, I'll start off with a small topological lemma to abstract out that part of the argument:



                            Lemma: Suppose $A subseteq [0, 1]$ satisfies $0 in A$, $A$ is closed under the left half-interval topology, and $A$ is open under the right half-interval topology. Then $A = [0, 1]$.



                            Proof (outline): Let $B := { x in [0, 1] mid [0, x] subseteq A }$ and $c := sup B$ (using the fact that $0 in B$ so $B$ is nonempty) and suppose that $c < 1$. Then from the condition that $A$ is closed under the left half-interval topology, we can conclude that $c in B$. On the other hand, the fact that $A$ is open under the right half-interval topology will now give a contradiction. Therefore, $c = 1$; and again, from the condition that $A$ is closed under the left half-interval topology, this will imply that $1 in B$, so $A = [0, 1]$ as desired. $square$



                            Now, to use this: to illustrate the essential points of the argument, I will restrict to the case where the domain of $f$ is $[0, 1]$ and $f(0) = 0$. The more general case should be straightforward to prove in a similar manner. Now, fix any $epsilon > 0$, and let $A_epsilon := { x in [0, 1] mid |f(x)| le epsilon x }$. Then $0 in A_epsilon$; and it should be straightforward to see that $A_epsilon$ is closed in the usual topology, and therefore also closed in the left half-interval topology. On the other hand, suppose $x in A_epsilon$; then from the hypothesis $f'(x) = 0$, we see that there exists $delta > 0$ such that whenever $0 < |y-x| < delta$, then $left| frac{f(y) - f(x)}{y - x} right| < epsilon$. Then, if in addition $y > x$, it follows that $|f(y)| le |f(x)| + |f(y) - f(x)| < epsilon x + epsilon |y-x| = epsilon y$; this shows that $A_epsilon$ is open in the right half-interval topology. Now from the lemma, it follows that $A_epsilon = [0, 1]$.



                            In summary, we have shown that $forall epsilon > 0, forall x in [0, 1], |f(x)| le epsilon x$. We may now interchange the two quantifiers to conclude $forall x in [0, 1], forall epsilon > 0, |f(x)| le epsilon x$. However, this easily implies $forall x in [0, 1], f(x) = 0$.






                            share|cite|improve this answer























                              up vote
                              0
                              down vote










                              up vote
                              0
                              down vote









                              Here is an alternate proof which bypasses the Mean Value Theorem, and which I hope might be more conducive to visualization. First, I'll start off with a small topological lemma to abstract out that part of the argument:



                              Lemma: Suppose $A subseteq [0, 1]$ satisfies $0 in A$, $A$ is closed under the left half-interval topology, and $A$ is open under the right half-interval topology. Then $A = [0, 1]$.



                              Proof (outline): Let $B := { x in [0, 1] mid [0, x] subseteq A }$ and $c := sup B$ (using the fact that $0 in B$ so $B$ is nonempty) and suppose that $c < 1$. Then from the condition that $A$ is closed under the left half-interval topology, we can conclude that $c in B$. On the other hand, the fact that $A$ is open under the right half-interval topology will now give a contradiction. Therefore, $c = 1$; and again, from the condition that $A$ is closed under the left half-interval topology, this will imply that $1 in B$, so $A = [0, 1]$ as desired. $square$



                              Now, to use this: to illustrate the essential points of the argument, I will restrict to the case where the domain of $f$ is $[0, 1]$ and $f(0) = 0$. The more general case should be straightforward to prove in a similar manner. Now, fix any $epsilon > 0$, and let $A_epsilon := { x in [0, 1] mid |f(x)| le epsilon x }$. Then $0 in A_epsilon$; and it should be straightforward to see that $A_epsilon$ is closed in the usual topology, and therefore also closed in the left half-interval topology. On the other hand, suppose $x in A_epsilon$; then from the hypothesis $f'(x) = 0$, we see that there exists $delta > 0$ such that whenever $0 < |y-x| < delta$, then $left| frac{f(y) - f(x)}{y - x} right| < epsilon$. Then, if in addition $y > x$, it follows that $|f(y)| le |f(x)| + |f(y) - f(x)| < epsilon x + epsilon |y-x| = epsilon y$; this shows that $A_epsilon$ is open in the right half-interval topology. Now from the lemma, it follows that $A_epsilon = [0, 1]$.



                              In summary, we have shown that $forall epsilon > 0, forall x in [0, 1], |f(x)| le epsilon x$. We may now interchange the two quantifiers to conclude $forall x in [0, 1], forall epsilon > 0, |f(x)| le epsilon x$. However, this easily implies $forall x in [0, 1], f(x) = 0$.






                              share|cite|improve this answer












                              Here is an alternate proof which bypasses the Mean Value Theorem, and which I hope might be more conducive to visualization. First, I'll start off with a small topological lemma to abstract out that part of the argument:



                              Lemma: Suppose $A subseteq [0, 1]$ satisfies $0 in A$, $A$ is closed under the left half-interval topology, and $A$ is open under the right half-interval topology. Then $A = [0, 1]$.



                              Proof (outline): Let $B := { x in [0, 1] mid [0, x] subseteq A }$ and $c := sup B$ (using the fact that $0 in B$ so $B$ is nonempty) and suppose that $c < 1$. Then from the condition that $A$ is closed under the left half-interval topology, we can conclude that $c in B$. On the other hand, the fact that $A$ is open under the right half-interval topology will now give a contradiction. Therefore, $c = 1$; and again, from the condition that $A$ is closed under the left half-interval topology, this will imply that $1 in B$, so $A = [0, 1]$ as desired. $square$



                              Now, to use this: to illustrate the essential points of the argument, I will restrict to the case where the domain of $f$ is $[0, 1]$ and $f(0) = 0$. The more general case should be straightforward to prove in a similar manner. Now, fix any $epsilon > 0$, and let $A_epsilon := { x in [0, 1] mid |f(x)| le epsilon x }$. Then $0 in A_epsilon$; and it should be straightforward to see that $A_epsilon$ is closed in the usual topology, and therefore also closed in the left half-interval topology. On the other hand, suppose $x in A_epsilon$; then from the hypothesis $f'(x) = 0$, we see that there exists $delta > 0$ such that whenever $0 < |y-x| < delta$, then $left| frac{f(y) - f(x)}{y - x} right| < epsilon$. Then, if in addition $y > x$, it follows that $|f(y)| le |f(x)| + |f(y) - f(x)| < epsilon x + epsilon |y-x| = epsilon y$; this shows that $A_epsilon$ is open in the right half-interval topology. Now from the lemma, it follows that $A_epsilon = [0, 1]$.



                              In summary, we have shown that $forall epsilon > 0, forall x in [0, 1], |f(x)| le epsilon x$. We may now interchange the two quantifiers to conclude $forall x in [0, 1], forall epsilon > 0, |f(x)| le epsilon x$. However, this easily implies $forall x in [0, 1], f(x) = 0$.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Nov 30 at 1:40









                              Daniel Schepler

                              8,2361618




                              8,2361618






















                                  up vote
                                  -1
                                  down vote













                                  Well, $f'(x)$ gives the rate of change of $f$ at a point $x$. If $f'(x)=0 forall x in (a,b)$, then $f(x)$ isn't changing, i.e. it is constant, on $(a,b)$.



                                  That's the idea, you just need to show it mathematically.






                                  share|cite|improve this answer

























                                    up vote
                                    -1
                                    down vote













                                    Well, $f'(x)$ gives the rate of change of $f$ at a point $x$. If $f'(x)=0 forall x in (a,b)$, then $f(x)$ isn't changing, i.e. it is constant, on $(a,b)$.



                                    That's the idea, you just need to show it mathematically.






                                    share|cite|improve this answer























                                      up vote
                                      -1
                                      down vote










                                      up vote
                                      -1
                                      down vote









                                      Well, $f'(x)$ gives the rate of change of $f$ at a point $x$. If $f'(x)=0 forall x in (a,b)$, then $f(x)$ isn't changing, i.e. it is constant, on $(a,b)$.



                                      That's the idea, you just need to show it mathematically.






                                      share|cite|improve this answer












                                      Well, $f'(x)$ gives the rate of change of $f$ at a point $x$. If $f'(x)=0 forall x in (a,b)$, then $f(x)$ isn't changing, i.e. it is constant, on $(a,b)$.



                                      That's the idea, you just need to show it mathematically.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Nov 29 at 4:25









                                      Rhys Hughes

                                      4,7081327




                                      4,7081327






























                                          draft saved

                                          draft discarded




















































                                          Thanks for contributing an answer to Mathematics Stack Exchange!


                                          • Please be sure to answer the question. Provide details and share your research!

                                          But avoid



                                          • Asking for help, clarification, or responding to other answers.

                                          • Making statements based on opinion; back them up with references or personal experience.


                                          Use MathJax to format equations. MathJax reference.


                                          To learn more, see our tips on writing great answers.





                                          Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                                          Please pay close attention to the following guidance:


                                          • Please be sure to answer the question. Provide details and share your research!

                                          But avoid



                                          • Asking for help, clarification, or responding to other answers.

                                          • Making statements based on opinion; back them up with references or personal experience.


                                          To learn more, see our tips on writing great answers.




                                          draft saved


                                          draft discarded














                                          StackExchange.ready(
                                          function () {
                                          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3018160%2fproving-a-continuous-function-is-a-constant%23new-answer', 'question_page');
                                          }
                                          );

                                          Post as a guest















                                          Required, but never shown





















































                                          Required, but never shown














                                          Required, but never shown












                                          Required, but never shown







                                          Required, but never shown

































                                          Required, but never shown














                                          Required, but never shown












                                          Required, but never shown







                                          Required, but never shown







                                          Popular posts from this blog

                                          Berounka

                                          Fiat S.p.A.

                                          Type 'String' is not a subtype of type 'int' of 'index'