# Characterization of strong minimums with slices.

I am doing a proof of a Lemma that isn't in a book.

Let $X$ a Banach space and $\emptyset\not=S\subset X$ closed of $X$. Let $f$ be a lower semicontinuous function bounded below in $S$.

I have that $\displaystyle\lim_{\alpha\to 0^+} \mathop{diameter}(Sl(-f,S,\alpha))=0$, where $Sl(-f,S,\alpha)=\{x\in S\;:\, f(x)<\displaystyle\inf_{z\in S}\{f(z)\}+\alpha\}$.

I need to prove that:

$\exists x\in S$ such that $f(s)=\inf_{z\in S}\{f(z)\}$ and given a sequence $(x_n)\subset S$ such that $f(x_n)\to f(x)$, then $\|x_n-x\|\to 0$.

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It is not true (perhaps you are missing a condition?)

Define the function $f:\mathbb{R}\rightarrow\mathbb{R}$ as follows: $$f(x) = \left\{ \begin{array}{lr} e^x & : x <0\\ |x-1| & : x \geq 0 \end{array}\right.$$

Then $f$ is continuous, bounded below, $S=\mathbb{R}$ is closed, $\inf_{x\in S} f(x) = 0$, and $x=1$ is the unique minimizer.

Now take the sequence $x_n = -n$. Then $f(x_n) \rightarrow 0$, but $x_n$ clearly does not converge.

Let $L_{\beta} = \{ x \in S | f(x) < \beta \}$, and $\underline f = \inf_{x\in S} f(x)$.

The collection of sets $L_{\underline{f}+\frac{1}{n}}$ is nested, non-empty and, by assumption, the diameter shrinks to $0$. If $x_n \in L_{\underline{f}+\frac{1}{n}}$, then it is not hard to see that $x_n$ is Cauchy, and hence converges to some $\hat{x} \in S$. Furthermore, we have $\underline{f} \leq f(x_n) < \underline{f}+\frac{1}{n}$, and since $x_n \rightarrow \hat{x}$ and $f$ is lower semi-continuous, we have $f(\hat{x}) \leq \liminf_{n \rightarrow \infty} f(x_n)$. Consequently, $f$ has a minimum on $S$ at $\hat{x}$.

Furthermore, if $f(\hat{x}) = f(y)$, with $y \in S$, the 'shrinking diameter' condition shows that $y=\hat{x}$, hence $\hat{x}$ is the unique minimizer of $f$ in $S$.

Now suppose $f(x_n) \rightarrow \underline{f} = f(\hat{x})$. Let $\epsilon > 0$. Then there exists a $\delta>0$ such that $\mathbb{diam} L_{\underline{f}+\delta} < \epsilon$, and a $N$ such that if $n>N$, then $x_n \in L_{\underline{f}+\delta}$. It follows that $x_n$ is Cauchy and converges to some element $y \in S$. As above, it follows that $f(y) = f(\hat{x})$ and hence that $y=\hat{x}$.

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There is only one example. I am defining $f$ on $\mathbb{R}$. Clearly $f(x) \geq 0$, $\forall x$, and there is exactly one point at which $f(x) = 0$, that is, $x=1$. It is a counterexample to your 'lemma'. – copper.hat May 7 '12 at 1:39
Ok, I edit it, better. – Shrek May 7 '12 at 15:31
That is more than an edit, you have completely changed the question. – copper.hat May 7 '12 at 16:28
In Banach spaces I think that a set it isn't compact if it is bounded and closed :S – Shrek May 22 '12 at 2:00
Ah yes, that was sloppy of me. I have modified the proof so it does not need compactness. – copper.hat May 22 '12 at 5:24