Let
- $V,W$ be $\mathbb R$-Banach spaces
- $\Omega\subseteq V$ be open
- $f:\Omega\to W$
It's easy to see that if $f$ is differentiable at $x_0\in\Omega$, then there is some $\delta>0$ with $$\left\|f(x)-f(x_0)\right\|_W\le(1+\left\|{\rm D}f(x_0)\right\|_{\mathfrak L(V,\:W)})\left\|x-x_0\right\|_V\;\;\;\text{for all }x\in B_\delta(x_0)\tag 1$$ where $B_\delta(x_0)$ is the open ball in $\Omega$ around $x_0$ with radius $\delta$, i.e. $f$ is locally Lipschitz continuous at $x_0$.
Now suppose that $f$ is differentiable (at each point) and ${\rm D}f$ is bounded. I would like to show that $f$ must be Lipschitz continuous, i.e. there is some $C\ge 0$ with $$\left\|f(x_1)-f(x_2)\right\|_W\le C\left\|x_1-x_2\right\|_V\;\;\;\text{for all }x_1,x_2\in\Omega.\tag 2$$
Let $x_1,x_2\in\Omega$. Since ${\rm D}f$ is bounded, there is some $\tilde C\ge 0$ with $$\left\|{\rm D}f(x_0)\right\|_{\mathfrak L(V,\:W)}\le\tilde C\;\;\;\text{for all }x_0\in V\tag 3$$ and hence we obtain the existence of $\delta_i>0$ with $$\left\|f(x)-f(x_i)\right\|_W\le(1+\tilde C)\left\|x-x_i\right\|_V\;\;\;\text{for all }x\in B_{\delta_i}(x_i)\tag 4$$ from $(1)$ for $i=1,2$.
However, in the case $B_{\delta_1}(x_1)\cap B_{\delta_2}(x_2)=\emptyset$ we cannot immediately conclude $(2)$ from $(4)$. So, the question is: How can we prove $(2)$?
The is called the mean value theorem (for differentiable functions in a Banach space). One version of the statement is that if $f$ is differentiable on a convex open set $U$, and $\|Df\|\leq M$ then $$ \forall x,y\in U : \|f(x) - f(y)\| \leq M \|x-y\|$$
A typical proof goes as follows: Let $\epsilon>0$ and consider the set $E$ of values $t\in [0,1]$ for which $$\forall \ 0\leq s\leq t \ \ \Rightarrow \ \|f(y+s(x-y))-f(y)\| \leq \epsilon+ s(M+\epsilon) \|x-y\|$$ Then $0\in E$ so $a=\sup E$ exists. The above inequality holds for $s=a$ by continuity of $f$. Suppose that $a<1$ then there is a squence $s_n\rightarrow a^+$ for which the inequality fails. Taking limits we must have: $$ \|f(y+a(x-y))-f(y)\| = \epsilon+ a(M+\epsilon) \|x-y\|$$ As $f$ is differentiable at $u=y+a(x-y)$ there is $\delta>0$ so that $$ \|f(u+s(x-y))-f(u)\| \leq (M+\epsilon)s $$ whenever $0\leq s\leq \delta$. Adding this to the previous we see by the triangular inequality that $a+\delta\in E$, contrary to the definition of $a$. Therefore, $a=1\in E$. Now, letting $\epsilon\rightarrow 0$ one obtains the claim.