Smooth convex functions Let $E\subseteq\mathbb R^d$ be a convex set, $\beta\geq 0$ be a given real number and $f:E\to\mathbb R$ be a convex and differentiable function satisfying:
$$f(y)\leq f(x)+\nabla f(x)^\top (y-x) +\frac{\beta}{2}\|x-y\|_2^2, \quad \forall x,y\in E.$$
Show that $\nabla f$ is $\beta$-Lipschitz, i.e.,
$$\|\nabla f(x)-\nabla f(y)\|_2\leq\beta\|x-y\|_2, \quad \forall x,y\in E.$$
Edit: I am interested in the case when $E$ can be any convex subset of $\mathbb R^d$. When $E=\mathbb R^d$, the claim can be proven as proposed by the answer below. However, in many convex optimisation textbooks, the claim is stated for general domains $E$ without proof.
 A: Let $E$ be open, convex and assume that $f$ is convex, differentiable and satisfies your inequality.
It is easy to check that $\nabla f$ is continuous:

Let $(x_n) \subset E$ be a sequence with $x_n \to x \in E$. Since $\nabla f(x_n)$ is bounded ($f$ is locally Lipschitz), we have $\nabla f(x_{n_k}) \to g$ for a subsequence. A standard argument shows that $g \in \partial f(x) = \{ \nabla f(x)\}$. A subsequence-subsequence argument shows that the entire sequence $(\nabla f(x_k))$ converges towards $\nabla f(x)$.

The answer by daw shows that we get
$$
\| \nabla f(y) - \nabla f(x) \| \le \beta \, \| y - x \|
$$
whenever $x,y \in E$ satisfy
$$
x + \frac1\beta \, \Big(\nabla f(y) - \nabla f(x)\Big),\;
y + \frac1\beta \, \Big(\nabla f(x) - \nabla f(y)\Big) \in E.$$
Now, since $E$ is open, we can check that there is $\varepsilon_x > 0$,
such that all $y \in B_{\varepsilon_x}(x)$ satisfy this condition.
Finally, let $x,y \in E$ be arbitrary. Since the segment $[x,y]$ is compact,
we can find finitely many $z_k \in [x,y]$, such that the corresponding balls cover $[x,y]$.
A repeated application of the triangle inequality shows the desired
$$
\| \nabla f(y) - \nabla f(x) \| \le \beta \, \| y - x \|.
$$
A: Let $x,y,z\in \mathbb R^d$. Then by the smoothness condition
$$
f(x+z) \le f(x) + \nabla f(x)^Tz + \frac\beta2 \|z\|_2^2
$$
and by convexity
$$
\nabla f(y)^T (x+z-y) \le f(x+z)-f(y).
$$
Adding both inequalities gives
$$
(\nabla f(y) - \nabla f(x))^Tz - \frac\beta2 \|z\|_2^2 \le f(x) - f(y) -\nabla f(y)^T(x-y).
$$
The left-hand side is maximal for $z= \frac1{\beta}(\nabla f(y) - \nabla f(x))$, which gives
$$
\frac1{2\beta} \|\nabla f(y) - \nabla f(x)\|_2^2 \le f(x) - f(y) -\nabla f(y)^T(x-y).
$$
The inequality is true if we exchange $x$ and $y$. Adding the resulting both inequalities yields to
$$
\frac1{\beta} \|\nabla f(y) - \nabla f(x)\|_2^2 \le (\nabla f(x)-\nabla f(y))^T(x-y).
$$
