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I'm doing Ex 3.32.5 in Brezis's book of Functional Analysis.

Let $(E, |\cdot|)$ be a uniformly convex Banach space. Let $\varphi: E \to (-\infty,+\infty]$ be a proper convex l.s.c. Then for every $x \in E$ and every integer $n \geq 1$, $$ \inf _{y \in E}\left\{n |x-y|^{2}+\varphi(y)\right\} $$ is achieved at some unique point, denoted by $y_{n}$.

My first step is to show that the map $y \mapsto |x-y|^{2}+\varphi(y)$ is bounded from below. However, I can show it in a very limited setting when $E = \mathbb R$ (and possibly $E = \mathbb R^d)$. Could you please shed me some lights?

Let $E := \mathbb R$. We fix some $x\in E$ and $n:=1$. Then there is an affine function $f: \mathbb R \to \mathbb R, y \mapsto ay+b$ with $a,b$ constants such that $\varphi(x) = f(x)$ and $\varphi(y) \ge f(y)$ for all $y\in E$. Then $$ |x-y|^2 + \varphi(y) \ge |x-y|^2 + ay+b =y^2+(a-2x)y+(x^2+b), $$ which is bounded from below in $y$.

Analyst
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  • You do not need to show boundedness from below. You need to show boundedness of a infimizing sequence. Use that a lsc convex functions is bounded from below by an affine function. – daw Mar 29 '22 at 19:57
  • @daw Could you give me some references that contain the proof of "a lsc convex functions is bounded from below by an affine function". I found one, but the proof there uses the inner product of Hilbert space. – Analyst Mar 30 '22 at 06:49
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    this should be contained in any book on convex analysis, for instance Barbu & Precupanu – daw Mar 30 '22 at 06:53

1 Answers1

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We fix $x \in E$ and $n \in \mathbb N$. It follows from this theorem that there exist $a\in \mathbb R$ and $f \in E'$ such that $\varphi(y) \ge f(y)+a$ for all $y\in E$. Then $$ n|x-y|^2 + \varphi(y) \ge n|x-y|^2 + f(y) + a \ge n|x-y|^2 - \|f\| |y| + a. $$

Clearly, the map $g:y \mapsto n|x-y|^2 - \|f\| |y| + a$ is bounded from below. Let $(y_m)$ be the minimizing sequence. Then $(y_m)$ is bounded. If not, $g(y_m) \to +\infty$, which is a contradiction.

Analyst
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