Consider the following statement:
$\textbf{Statement:} $ Let $X$ be a normed space, $S\subseteq X$ and $\bar{x}\in S.$ Let $f:X \to \Bbb R$ be locally Lipschitz and directionally differentiable at $\bar{x},$ that is, the limit $$\lim_{t\to 0^+}\frac{f(\bar{x}+tv)-f(\bar{x})}{t}=f'(\bar{x},v)$$ exists for all $v\in X.$ Finally, consider the problem $$(P)\;\min_{x\in S} f(x).$$ Then,
$$f'(\bar{x},v)>0 \; \forall\; v\in T(S,\bar{x})\setminus\{0\}\Longrightarrow \bar{x} \textrm{ is a strict local minimum of (P)}.$$ Here,
$T(S, \bar{x})= \{d\in X\;|\;\exists\; t_n\to 0^+,\exists \;d_n\to d: \bar{x}+t_nd_n\in S\}$ is the contingent cone of $S$ at $\bar{x}.$ $\Box$
I have a couple of questions about this statement:
- Is it true in any normed space?
I know it is true at least in finite dimensions. The proof is as follows: Let $X=\Bbb R^n.$ Assume the statement is not true. then you can find a sequence $\{x_n\}\subseteq S$ such that $x_n\to \bar{x}$ and $f(x_n)\leq f(\bar{x}).$ Consider the sequence $d_n=\frac{x_n-\bar{x}}{\|x_n-\bar{x}\|},$ and put $t_n= \|x_n-\bar{x}\|.$ Then $t_n\to 0^+$ and $x_n= \bar{x}+t_nd_n \in S.$
Since $d_n$ is in the unit sphere, which is $\textbf{compact},$ without loss of generality we can assume that $\{d_n\}$ is convergent to some $d\in \Bbb R^n.$ Then, we find that
$0\geq \frac{f(x_n)-f(\bar{x})}{t_n}= \frac{f(\bar{x}+t_nd_n)-f(\bar{x})}{t_n}\to f'(\bar{x},d).$ Note that in the last part we use the fact that $f$ is locally Lipschitz to guarantee the convergence to the directional derivative.
hence, we obtained a contradiction and the statement must be true.
However, when $X$ is infinite dimensional, we can not go around this lines because of the compactness assumption. Which brings me to the second question:
- If the statement is not true, can you provide a counterexample?
and finally, but most important,
- Under what assumptions on $X$(Banach, reflexivity, etc) can we guarantee the validity of the statement?
A reference is also very helpful. Thanks in advance to all!
$\textbf{EDIT:}$ I found some paper on sufficient first-order optimality conditions. It turns out that the statement is true if we further assume that there exists $\alpha >0$ such that
$$(i) \;\;\forall \;v\in T(S,\bar{x}): f'(\bar{x}, v)\geq \alpha\|v\|$$ and
$$(ii) \;\; d(x-\bar{x},T(S;\bar{x}))=o(\|x-\bar{x}\|).$$ Condition $(ii)$ is always true in finite dimmensions. Furhermore, if $T(S,\bar{x})$ has a weakly compact base(which is true in finite dimmensions), then $(i)$ is equivalent to $$\forall \;v\in T(S,\bar{x}): f'(\bar{x}, v)>0,$$ which is the original statement.
The claim is not true. Consider the following example: $X=L^2(0,1)$, $$ f(x) = -\frac12 \|x\|_{L^2}^2, $$ $$ S= \{x: -1\le x\le 1\text{ a.e. }\}. $$ Then $f$ is locally Lipschitz and Frechet differentiable. The point $x^*\equiv -1$ is a global minimum of $f$ on $S$. The tangent cone is $$ T_S(x^*) = \{ h: \ h\ge 0 \text{ a.e. } \}. $$ Since $f'(x)h = -(x,h)_{L^2}$, this shows $$ f'(x^*)h = \int_0^1 h(t)dt >0 \quad \forall h\in T_S(x^*)\setminus \{0\}. $$ However, $x^*$ is not a strict minimum. Consider the sequence $x_\epsilon:=\chi_{(0,\epsilon)} - \chi_{(\epsilon,1)}$. Then $x_\epsilon\to x^*$ in $L^2$ and $f(x^*)=f(x_\epsilon)=-1$.
This example shows that there is not much hope for the general case: The space is a separable Hilbert space. The objective functional is Frechet. The constraint set is convex (hence replacing contingent cone by a generalized tangent cone will not help).
I fear the only way to prove this result in a general setting is to assume convexity of $f$ or something similarly strong.