Theorem 1   Let $X,Y$ be Banach spaces and let $A$ be a convex (see convex set), open subset of $X$ Let $f\colon \overline{A}\to Y$ be a function which is continuous in $\overline A$ and differentiable in $A$ Then $f$ is lipschitz continuous on $\overline A$ if and only if the derivative $Df$ is bounded on $A$ i.e. $$ \sup_{x\in A} \Vert Df(x) \Vert < +\infty. $$

Proof. Suppose that $f$ is lipschitz continuous: $$ \Vert f(x) - f(y)\Vert \le L \Vert x - y\Vert. $$ Then given any $x\in A$ and any $v\in X$ for all small $h\in \mathbb R$ we have $$ \Vert \frac{f(x+hv)-f(x)}{h}\Vert \le L. $$ Hence, passing to the limit $h\to 0$ it must hold $\Vert Df(x)\Vert\le L$

On the other hand suppose that $Df$ is bounded on $A$ $$ \Vert Df(x)\Vert \le L,\qquad \forall x \in A. $$ Given any two points $x,y\in \overline A$ and given any $\alpha\in Y^*$ consider the function $G:[0,1]\to \mathbb R$ $$ G(t) = \langle \alpha, f((1-t)x +t y)\rangle. $$ For $t\in (0,1)$ it holds $$ G'(t) = \langle \alpha, Df((1-t)x+ty)[y-x]\rangle $$ and hence $$ \vert G'(t)\vert \le L \Vert \alpha\Vert\, \Vert y-x\Vert. $$ Applying Lagrange mean-value theorem to $G$ we know that there exists $\xi\in(0,1)$ such that $$ \vert \langle \alpha, f(y)-f(x)\rangle\vert = \vert G(1)-G(0)\vert = \vert G'(\xi)\vert \le \Vert \alpha\Vert L \Vert y-x\Vert $$ and since this is true for all $\alpha\in Y^*$ we get $$ \Vert f(y)-f(x) \Vert \le L \Vert y-x\Vert $$ which is the desired claim.