Htykuut

Full question:

Suppose $f:Uto mathbb{R}^m$ is differentiable and $U$ is convex. If there exists an $M>0$ such that $vert Df(x)vertleq M$ $forall xin U$ Show that $vert f(b)-f(a)vert leq Mvert b-a vert$.

My proof:

Fix some arbitrary $a,bin U$
Since $f:mathbb{R}^ntomathbb{R}^m$, each component function $f_i$ is a function from $mathbb{R}^ntomathbb{R}$
Define the function $gamma:mathbb{R}to U$ by $gamma (t)=vec{a}+t(vec{b}-vec{a})$

Let $h(t):=f_icircgamma (t)$, where $f_i$ is any component function of $f$. We can do this because U is convex and contains the straight line from a to b. Since $f$ is differentiable we know each $f_i$ is differentiable.
And $h(t):mathbb{R}tomathbb{R}$

Apply FTC for $h(1)-h(0)$ we get:
$h(1)-h(0)=f_i(b)-f_i(a)=int_0^1frac{d}{dx}f_i(gamma(t))dt=int_0^1nabla f_i(vec{a}+t(vec{b}-vec{a}))cdot (vec{b}-vec{a})dt$

then $vert f_i(vec{b})-f_i(vec{a})vert=vertint_0^1nabla f_i(vec{a}+t(vec{b}-vec{a}))cdot(vec{b}-vec{a})dtvert$

$leqint_0^1vertnabla f_i(vec{a}+t(vec{b}-vec{a}))cdot (vec{b}-vec{a})vert dt$ $leq int_0^1 vertnabla f_i(vec{a}+t(vec{b}-vec{a}))vertvert(b-a)vert dt$

By assumption $vertnabla f_i(x)vertleq M$ thus
$vert f_i(vec{b})-f_i(vec{a})vertleq Mvert(vec{b}-vec{a})vertint_0^1dt=Mvert(vec{b}-vec{a})vert$

then for $f:mathbb{R}^ntomathbb{R}^m$
$vertvert f(vec{b})-f(vec{a})vertvert=sqrt{sum_{i=1}^m(f_i(vec{b})-f_i(vec{a}))^2}leqsqrt{M^2vert(vec{b}-vec{a})vert^2(m)}=sqrt{m}Mvert(vec{b}-vec{a})vert$

The problem here is that I get this $sqrt{m}$, the solution apparently should be that you can show its bounded just by $Mvert b-a vert$.

Consider the case of a vector $v$ where $|v| leq 1$. Then each $|v_i| leq 1$ and so $|v| = sqrt{v_1^2+cdots+v_n^2} leq sqrt{ncdot 1} = sqrt{n}$. True, but we've lost some precision! That's where your square root came in and you can find exactly in your proof where that happened. So a different approach, using the total derivative will be necessary.

Adapted from Sternberg's Advanced Calculus you can argue as follows. Call $h(t) = f(gamma(t))$ so $h:[0,1]rightarrow mathbb{R}^m$ and by the chain rule $|h'(t)| leq M|b-a|$. Fix $epsilon > 0$. Consider the "good set" $$G = {tin[0,1]:; |h(t)-h(0)| leq (M|b-a|+epsilon)t + epsilon }$$ If we show that $1in G$ then since $epsilon$ is arbitrary, the result follows (You'll see why that multiplicative $epsilon$ appears in a minute).

Firstly note by continuity of the map $tmapsto |h(t)-h(0)|-(M|b-a|+epsilon)t$ at zero, the set $G$ must contain some interval to the right of $0$. Let $t^* = sup G$ and again by continuity it is seen that $t^* in G$. If (for contradiction) $t^* neq 1$ then by definition of the derivative $h'(t^*)$:

$$
frac{|h(t)-h(t^*)|}{t-t^*} leq M|b-a| + epsilon
$$
for all $tin (t^*,t^*+delta)$ for some $delta>0$. Then for such $t$:
begin{align*}
|h(t)-h(0)| &leq |h(t)-h(t^*)|+|h(t^*)-h(0)| \
&leq (M|b-a|+epsilon)(t-t^*) + (M|b-a|+epsilon)t^* +epsilon\
&=(M|b-a|+epsilon)t+epsilon
end{align*}
But that means $t$ belongs to $G$, a contradiction since $t>t^*$.

The inequality $|f(b)-f(a) | leq M|b-a|$ is equivalent to $|g(b)-g(a) | leq M|b-a|$ for every vector $c$ of norm $1$ where $g(x)=sum f_i(x)c_i$. This reduces the proof too the case $m=1$ and your argument works now.