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Let $(Y,\|\cdot\|)$ a Banach space and $b\colon Y\to \mathbb{R}$ a nonnegative convex function such that, for some $\mathcal{E}>0$, the set $\{y\in Y\,:\, b(y)<\mathcal{E}\}$ is nonempty and bounded.

I need prove that exist $z\in Y$, $c>0$ and $C>0$ such that $b(y)-b(z)\geq c\|y-z\|$ for $y\in Y\backslash B_C(z)$.

Note: $B_C(z):=\{y\in Y\,:\, \|y-z\|\leq C\}$.

Thanks in advance.

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up vote 2 down vote accepted

Let $\epsilon_0$ be the fixed positive constant such that $F_0 := \{ y \mid b(y) < \epsilon_0\}$ is nonempty and bounded. Fix $z\in F_0$. Then we have that if $y \in Y \setminus F_0$, $b(y) - b(z) \geq \epsilon_0 - b(z) =: \delta_0 > 0$.

By assumption $F_0$ is bounded, hence there exists $C > 0$ such that $B_C(z) \supset F_0$. (Here I take $B_C(z)$ to be the open ball; a simple modification of the following argument also allows for closed balls.)

Let $c = \frac{\delta_0}{2C}$. Now we verify this choice of $C$ and $c$ is good.

Given $y \in Y\setminus B_C(z)$, let $\tilde{y} = \frac{C}{\|y-z\|} (y-z) + z$. By convexity of the function $b$ we have that

$$ \frac{C}{\|y-z\|} (b(y) - b(z)) + b(z) \geq b(\tilde{y}) \geq b(z) + \delta_0 $$

Hence we have that

$$ b(y) - b(z) \geq \frac{\delta_0}{C} \|y-z\| > \frac{\delta_0}{2C} \|y-z\| = c \| y - z\| $$

as claimed.

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