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If $f : E \to F$ is a continuous map of Banach spaces, with bounded Fréchet derivative. Then

$x_0,x_1 \in E\Rightarrow \|f(x_1) − f(x_0)\| ≤ M\|x_1 − x_0\|$ where $M = \sup \|f'(x)\|.$

The most efficient way to prove this, as far as I know, is to apply Hahn-Banach. Alternatively, one starts with $g(t)= f(x_0 + t(x_1 − x_0));\ 0\le t\le 1$ and reduces the problem to the real case. But, of course, the Hahn-Banach theorem is a (relatively) big gun, and to do the integral if one uses $g$, then integration in Banach spaces must be dealt with (via regulated functions, for example.)

The only elementary proof I have seen uses the dot product (Apostol, Rudin), but then of course, one must assume an inner product.

Below I sketch a very basic proof, but it required more effort than I expected, so my question is: can one get it cheaper, using only elementary means (very basic facts about normed spaces and the definition of derivative)?

Assume for convenience that $x_0=f(x_0)=0$ and consider the segment $\{tx_1:\ 0\le t\le 1\}.$

We have $\|f(x)\|\le M\|x\|+\|r(x)\|\cdot\|x\|$ where $\frac{\|r(x)\|}{\|x\|}\to 0$ as $x\to 0.$

So, if $\epsilon>0,$ we can choose $\delta>0$ such that $t\le \min \left(1,\frac{\delta}{\|x_1\|}\right)\Rightarrow \|f(tx_1)\| ≤ (M + \epsilon)t\|x_1\|$. To prove the claim, it will suffice to prove that $t=1$ satisfies this last inequality, so to that end,

let $\tau=\sup\{r>0:\|f(tx_1)\| ≤ (M + \epsilon)t\|x_1\|\text{for all}\ t\in [0,r]\}$, so that $\tau\le 1$ and $\tau\in \{r>0:\|f(tx_1)\| ≤ (M + \epsilon)t\|x_1\|\text{for all}\ t\in [0,r]\}$ (because $f$ is continuous.)

Now, toward a contradiction, suppose that $\tau<1$ and choose $c>0$ and small enough so that $(\tau+c)x_1$ is on the segment.

Now, there is a $\delta' > 0$ such that

$\|f(x) − f(\tau x_1)\| \le (M + \epsilon)||x − \tau x_1\|<\epsilon$ whenever $\|x − \tau x_1\| <\delta'.$

So, since $\|(\tau + c)x_1 − \tau x_1\| = c\|x_1\|<\delta' $ if $c$ is small enough, we have, with $x=(\tau + c)x_1$,

$\|f((\tau+c) x_1)-f(\tau x_1)\| \le (M+\epsilon)\|(\tau+c)x_1-\tau x_1\|=(M+\epsilon)\|cx_1\|$

and so finally,

$f((\tau+c)x_1)\le (M+\epsilon)\|cx_1\|+\|f(\tau x_1)\|\le $

$(M+\epsilon)\|cx_1\|+ (M+\epsilon)\|\tau x_1\|=(M+\epsilon)(c+\tau)\|x_1\|$,

which contradicts the fact that $\tau$ is a supremum.

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  • $\begingroup$ This is Theorem 1.1.1 in Hörmander's The Analysis of Partial Differential Operators. He needs 8 lines for the proof. $\endgroup$ – Jochen Mar 29 at 11:13
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Here is another proof. At the core, the arguments are similar to yours (but with fewer $\delta$s).

Assume that there are $x_1,y_1\in E$ and $\epsilon>0$ such that $$ \|f(y_1)-f(x_1) \| > (M+\epsilon) \|y_1-x_1\|. $$ By triangle inequality, one of the following inequalities are satisfied: $$ \|f(y_1)-f(\frac{x_1+y_2}2) \| > (M+\epsilon) \|y_1-\frac{x_1+y_2}2\|, $$ $$ \|f(\frac{x_1+y_2}2) - f(x_1)\| > (M+\epsilon) \|\frac{x_1+y_2}2-x_1\|. $$ Inductively, we can construct elements $(x_n,y_n)$ on the line segment $[x_1,y_1]$ with $\|x_n-y_n\| = 2^{1-n} \|x_1-y_1\|$, $[x_{n+1},y_{n+1}]\subseteq[x_n,y_n]$ and $$ \|f(y_n)-f(x_n) \| > (M+\epsilon) \|y_n-x_n\|. $$ Then $x_n,y_n\to x$ for some $x\in [x_n,y_n]$. And $$ \|f(y_n)-f(x_n) \| \le \|f(y_n)-f(x)\| + \|f(x_n)-f(x)\|. $$ By Frechet differentiability of $f$ at $x$ there we get for $n$ large enough $$\begin{split} \|f(y_n)-f(x_n) \| &\le \|f(y_n)-f(x)\| + \|f(x_n)-f(x)\| \\ &\le (M+\epsilon/2) (\|y_n-x\| + \|x-x_n\|) \\& = (M+\epsilon/2)\|y_n-x_n\|, \end{split}$$ which is a contradiction.


Let me try a direct proof in the flavor of your post. The proof works just with Gateaux differentiability of $f$ (and does not need continuity). Let $x,h\in E$ be given. Let $\epsilon>0$. Define $$ I = \{ t\in [0,1]: \|f(x+th)-f(x)\| \le (M+\epsilon) t\|h\|\}. $$ Clearly, $0\in I$. Assume $[0,t]\subset I$ for some $t\ge0$. Then by Gateaux differentiability of $f$ at $x+th$, there is $s_0>0$ such that $$ \|f(x+(s+t)h)-f(x+th)\| \le (M+\epsilon) s\cdot \|h\| $$ for $s \in (0, s_0)$. By triangle inequality, $$ \|f(x+(s+t)h)-f(x)\| \le (M+\epsilon) (s+t)\cdot \|h\| \quad \forall s\in (0,s_0). $$ Hence $[0,t+s_0)\cap [0,1]\subset I$.

Now suppose $[0,t)\subset I$ for some $t>0$. By Gateaux differentiability of $f$ at $x+th$, there is $s_0>0$ such that $$ \|f(x+(t-s)h)-f(x+th)\| \le (M+\epsilon) s\cdot \|h\| $$ for $s \in (0, s_0)$. Taking $s\in (0,s_0$ such that $t-s\in I$, and using triangle inequality as above, it follows $t\in I$ and $[0,t]\subset I$.

By induction on the reals, $I=[0,1]$: Assume $I\ne[0,1]$. Let $t = \inf [0,1]\setminus I$. Then $t\ne0$ because there is $s_0>0$ such that $[0,s_0)\subset I$. By definition of inf, $[0,t)\subset I$. Using the steps above: $[0,t]\subset I$, $[0,t+s_0)\subset I$ for some $s_0>0$.

Since $\epsilon$ in the definition of $I$ was arbitrary, the claim follows.

Using these arguments, one can prove for Gateaux differentiable $f$ the standard mean value inequality $$ \|f(x+h)-f(x) \| \le \sup_{t\in [0,1)}\|f'(x+th)\| \|h\|, $$ with supremum taken over the half-open interval.

Can this be reduced to a supremum over the open interval $(0,1)$?

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  • $\begingroup$ You only need completeness of the (at most two-dimensional) space spanned by $x_1$ and $y_1$. $\endgroup$ – gerw Mar 29 at 7:58
  • $\begingroup$ @gerw so true....... $\endgroup$ – daw Mar 29 at 8:00

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