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Let consider a normed vector space $V$. I want to prove that

If $f:V\to \mathbb R$ is a convex function and if for some $x_0 \in V$ the function is bounded on a neighborhood $W$ of $x_0$, then there exists a neighborhood $U$ of $x_0$ such that $f$ is Lipschitz on $U$.

When I say neighborhood of $x_0$ I mean an open set that contains $x_0$ WLOG the neighborhood clearly can be considered to be a ball with center $x_0$.

I don't know what can I do, because I can compute directly $f(x)-f(y)$ because the convexity only works with positive scalars. So I'm a little confused, please help me!

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up vote 2 down vote accepted

Choose $r > 0$ such that $f$ is bounded on $\bar U_{2r}(x_0)$. We will prove that $f$ is Lipschitz on $\bar U_r(x_0)$. So let $x,y \in \bar U_r(x_0)$, choose $z \in \partial U_{2r}(x_0)$ such that the ray $x + [0,\infty) \cdot (y-x)$ intersects $\partial U_{2r}(x_0)$ in $z$. Let $\lambda := \def\norm#1{\left\|#1\right\|}{\norm{y-x}}\bigm/{\norm{z-x}}$. By convexity, we have \begin{align*} f(y) - f(x) &\le \lambda \cdot \bigl(f(z) - f(x)\bigr)\\ &= \frac{f(z)-f(x)}{\norm{z-x}}\cdot \norm{y-x} \end{align*} Now, as $\norm{z-x} \ge r$, we get $$ f(y) - f(x) \le \frac{2\norm{f}_{\infty, \bar U_{2r}(x_0)}}r \cdot \norm{y-x} $$ Swapping the roles of $x$ and $y$, we get finally $$ \def\abs#1{\left|#1\right|} \abs{f(y) - f(x)} \le \frac{2\norm{f}_{\infty, \bar U_{2r}(x_0)}}r \cdot \norm{y-x}, \quad x,y \in \bar U_r(x_0). $$

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Very good answer! – Trafalgar Law Jun 14 '13 at 7:33

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