Help understanding norm in $mathcal{L}(X, mathbb{R}) $ and unbounded linear operator
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if $f in mathcal{L}(X, mathbb{R} ) $ is discontinuous at $0$ in $X$ , show that ${x_n} rightarrow 0 $ with $f(x_n)=1$ for all n.
Since $f$ is discontinuous at $0$ then $||f||=+ infty $, but I need to find ${x_n} rightarrow 0 $ with $f(x_n) = 1$
$forall n$ I don't get it here...
is $||f||=sup_{||x||=1} |f(x)|$ ?
I have find that $x_n= frac{1}{sum f(xn)} rightarrow 0$ if $f(x_n)=1$ for all n.
functional-analysis
asked Nov 24 at 20:28
Alexei
32
add a comment |
up vote
0
down vote
favorite
if $f in mathcal{L}(X, mathbb{R} ) $ is discontinuous at $0$ in $X$ , show that ${x_n} rightarrow 0 $ with $f(x_n)=1$ for all n.
Since $f$ is discontinuous at $0$ then $||f||=+ infty $, but I need to find ${x_n} rightarrow 0 $ with $f(x_n) = 1$
$forall n$ I don't get it here...
is $||f||=sup_{||x||=1} |f(x)|$ ?
I have find that $x_n= frac{1}{sum f(xn)} rightarrow 0$ if $f(x_n)=1$ for all n.
functional-analysis
asked Nov 24 at 20:28
Alexei
32
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
if $f in mathcal{L}(X, mathbb{R} ) $ is discontinuous at $0$ in $X$ , show that ${x_n} rightarrow 0 $ with $f(x_n)=1$ for all n.
Since $f$ is discontinuous at $0$ then $||f||=+ infty $, but I need to find ${x_n} rightarrow 0 $ with $f(x_n) = 1$
$forall n$ I don't get it here...
is $||f||=sup_{||x||=1} |f(x)|$ ?
I have find that $x_n= frac{1}{sum f(xn)} rightarrow 0$ if $f(x_n)=1$ for all n.
functional-analysis
asked Nov 24 at 20:28
Alexei
32
if $f in mathcal{L}(X, mathbb{R} ) $ is discontinuous at $0$ in $X$ , show that ${x_n} rightarrow 0 $ with $f(x_n)=1$ for all n.
Since $f$ is discontinuous at $0$ then $||f||=+ infty $, but I need to find ${x_n} rightarrow 0 $ with $f(x_n) = 1$
$forall n$ I don't get it here...
is $||f||=sup_{||x||=1} |f(x)|$ ?
I have find that $x_n= frac{1}{sum f(xn)} rightarrow 0$ if $f(x_n)=1$ for all n.
functional-analysis
functional-analysis
asked Nov 24 at 20:28
Alexei
32
asked Nov 24 at 20:28
Alexei
32
edited Nov 24 at 20:44
asked Nov 24 at 20:28
Alexei
32
asked Nov 24 at 20:28
Alexei
32
asked Nov 24 at 20:28
Alexei
32
32
add a comment |
add a comment |
1 Answer
1
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Since $f$ is an unbounded linear operator, we can find a sequence ${y_n}_n subseteq {yin Xmid lVert yrVert le 1}$ such that $|f(y_n)| rightarrow infty$. Let $x_n = frac{y_n}{f(y_n)}$, then $f(x_n) = 1$ and $lVert x_nrVert le frac 1{|f(y_n)|}$. Since $frac 1{|f(y_n)|} rightarrow 0$, $x_nrightarrow 0$.
answered Nov 24 at 20:47
Syuizen
924311
sorry I don't understand why f(x_n)=1 for all n but |f(y_n)| diverge
– Alexei
Nov 24 at 20:53
@Alexei $f$ is linear so $f(x_n) = (f(y_n))/f(y_n) = 1$.
– Syuizen
Nov 24 at 20:57
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Since $f$ is an unbounded linear operator, we can find a sequence ${y_n}_n subseteq {yin Xmid lVert yrVert le 1}$ such that $|f(y_n)| rightarrow infty$. Let $x_n = frac{y_n}{f(y_n)}$, then $f(x_n) = 1$ and $lVert x_nrVert le frac 1{|f(y_n)|}$. Since $frac 1{|f(y_n)|} rightarrow 0$, $x_nrightarrow 0$.
answered Nov 24 at 20:47
Syuizen
924311
sorry I don't understand why f(x_n)=1 for all n but |f(y_n)| diverge
– Alexei
Nov 24 at 20:53
@Alexei $f$ is linear so $f(x_n) = (f(y_n))/f(y_n) = 1$.
– Syuizen
Nov 24 at 20:57
add a comment |
up vote
0
down vote
accepted
Since $f$ is an unbounded linear operator, we can find a sequence ${y_n}_n subseteq {yin Xmid lVert yrVert le 1}$ such that $|f(y_n)| rightarrow infty$. Let $x_n = frac{y_n}{f(y_n)}$, then $f(x_n) = 1$ and $lVert x_nrVert le frac 1{|f(y_n)|}$. Since $frac 1{|f(y_n)|} rightarrow 0$, $x_nrightarrow 0$.
answered Nov 24 at 20:47
Syuizen
924311
sorry I don't understand why f(x_n)=1 for all n but |f(y_n)| diverge
– Alexei
Nov 24 at 20:53
@Alexei $f$ is linear so $f(x_n) = (f(y_n))/f(y_n) = 1$.
– Syuizen
Nov 24 at 20:57
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Since $f$ is an unbounded linear operator, we can find a sequence ${y_n}_n subseteq {yin Xmid lVert yrVert le 1}$ such that $|f(y_n)| rightarrow infty$. Let $x_n = frac{y_n}{f(y_n)}$, then $f(x_n) = 1$ and $lVert x_nrVert le frac 1{|f(y_n)|}$. Since $frac 1{|f(y_n)|} rightarrow 0$, $x_nrightarrow 0$.
answered Nov 24 at 20:47
Syuizen
924311
Since $f$ is an unbounded linear operator, we can find a sequence ${y_n}_n subseteq {yin Xmid lVert yrVert le 1}$ such that $|f(y_n)| rightarrow infty$. Let $x_n = frac{y_n}{f(y_n)}$, then $f(x_n) = 1$ and $lVert x_nrVert le frac 1{|f(y_n)|}$. Since $frac 1{|f(y_n)|} rightarrow 0$, $x_nrightarrow 0$.
answered Nov 24 at 20:47
Syuizen
924311
edited Nov 24 at 20:57
answered Nov 24 at 20:47
Syuizen
924311
answered Nov 24 at 20:47
Syuizen
924311
answered Nov 24 at 20:47
Syuizen
924311
924311
sorry I don't understand why f(x_n)=1 for all n but |f(y_n)| diverge
– Alexei
Nov 24 at 20:53
@Alexei $f$ is linear so $f(x_n) = (f(y_n))/f(y_n) = 1$.
– Syuizen
Nov 24 at 20:57
add a comment |
sorry I don't understand why f(x_n)=1 for all n but |f(y_n)| diverge
– Alexei
Nov 24 at 20:53
@Alexei $f$ is linear so $f(x_n) = (f(y_n))/f(y_n) = 1$.
– Syuizen
Nov 24 at 20:57
sorry I don't understand why f(x_n)=1 for all n but |f(y_n)| diverge
– Alexei
Nov 24 at 20:53
sorry I don't understand why f(x_n)=1 for all n but |f(y_n)| diverge
– Alexei
Nov 24 at 20:53
@Alexei $f$ is linear so $f(x_n) = (f(y_n))/f(y_n) = 1$.
– Syuizen
Nov 24 at 20:57
@Alexei $f$ is linear so $f(x_n) = (f(y_n))/f(y_n) = 1$.
– Syuizen
Nov 24 at 20:57
add a comment |
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