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I am trying to show the following statement:

Let $f,g:\mathbb{R}^n\to \mathbb{R}\cup \{\infty\}$ be two convex functions. Assume that there are constants $C_1,C_2>0,\alpha>1$ such that $f(x)\geq C_1|x|^\alpha+C_2$. Show that $f\Box g(x):=\inf_{y\in\mathbb{R}^n}[f(y)+g(x-y)]\not= -\infty$ for all $x.$

I formed a proof with an additional assumption that $g(x)<\infty\forall x$, then I can use the facts that $g$ is continuous on $\mathbb{R^n}$ and $g(y)\geq h(y),\forall y\in \mathbb{R}^n$ for some affine function $h$ since $g$ has a supporting plane.

I think since we are trying to prove $f\Box g(x)=\inf_{y\in\mathbb{R}^n}[f(y)+g(x-y)]\not= -\infty$, I am fine to add this assumption without loss of generality. But I don't know how to argue it rigorously.

Can I add this assumption? If not, could you help to provide another method or some references? Thanks!

John
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1 Answers1

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There are two cases to consider:

(i) Suppose $g$ is not proper, then since $-\infty$ is excluded from the range, the epigraph must be empty and so $g= \infty$. Then $f \Box g = \infty$.

(ii) If $g$ is proper, it majorises some affine functional (see, for example, Corollary 12.1.2 in Rockafellar's "Convex Analysis"), that is, there is some $c, h$ such that $g(x) \ge c + \langle h, x \rangle$ for all $x$.

Then (where I am taking $C_1 \le 1$ without loss of generality) \begin{eqnarray} f(y)+g(x-y) &\ge& C_2 +C_1 \|y\|^\alpha + c + \langle h, x-y \rangle \\ &\ge& (C_2+c+\langle h, x \rangle) + C_1\|y\|^\alpha - \|h\| \|y\| \\ &\ge& (C_2+c+\langle h, x \rangle) + \|y\| (\|y\|^{\alpha-1}-\|h\|) \end{eqnarray} Since the function $y \mapsto \|y\| (\|y\|^{\alpha-1}-\|h\|)$ is bounded below, we have the desired result (in fact, this shows that $f \Box g$ is bounded below by the affine function $x \mapsto c' + \langle h, x \rangle$ for some $c'$).

copper.hat
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