Ytukyg

Let $f:X rightarrow mathbb{R}$ be convex, $x in X$. Assume further that $f$ is continuous and finite in $x.$ Then it says that by Hahn-Banach there is $x^* in X^*$ with

begin{equation}
langle y - x, x^* rangle_{X times X^*} + f(x) leq f(y) forall y in X.
end{equation}

I'm trying to understand why this is true. As far as I know, by Hahn-Banach we can seperate the convex set given by the points above the graph of $f$ from the point $(x,f(x))$ by some hyperplane. More generally, for $A,B subset X$, $Acap B= emptyset$ convex sets with $A$ open there is $x^* in X^*$ and $gamma in mathbb{R}$ such that

begin{equation}
langle a,x^* rangle _{X times X^*} leq gamma leq langle b,x^* rangle _{X times X^*} forall ain A, b in B. (1)
end{equation}

Intuitively, the above estimate makes sense. The desired hyperplane is an affine function below the graph of $f$ which intersects with the graph in the point $(x,f(x))$. But I would like to understand how this follows from the formal statement $(1)$ in the Hahn-Banach Theorem. Thank you for any help.

Let me add another answer to address your modified question.

What we want to do is to use the Hahn-Banach separation theorem with the open set $(text{epi} f)^circ$ and the compact set ${(x_0,f(x_0)}$. Continuity at $x$ guarantee that $(text{epi} f)^circ$ is not empty, which is not the case if $f$ is a discontinuous linear functional.

Recall that the epigraph $text{epi} f:= {(y,r)in XtimesBbb R: rge f(y)}$ is convex, and that $(Xtimes Bbb R)^*simeq X^*times Bbb R$. Note also that for a convex set $C$ with nonempty interior, we have $overline C = overline{(C^circ)}$.

Let $x_0$ be a point in $X$, take a small $varepsilon>0$ then by continuity of $f$ we can find a neighborhood $V$ of $x_0$ such that $f(y)<f(x_0)+varepsilon/2$ for all $yin V$. This means that
$$
Vtimes (f(x_0)+varepsilon,+ infty) subset text{epi} f
$$
and hence $(text{epi} f)^circne emptyset$.

Now, let's apply the Hahn-Banach separation theorem to the open set $(text{epi} f)^circ$ and the compact set ${(x_0,f(x_0)}$ to get $(x^*,lambda)in X^*times Bbb R$ such that
$$
langle x_0,x^* rangle + lambda f(x_0) > langle y,x^* rangle + lambda r
$$
for all $(y,r)in (text{epi} f)^circ$.

The above argument with small enough $varepsilon$ shows that $(x_0,r)in (text{epi} f)^circ$ for all $r>f(x_0)$, substitute this into $(y,r)$ and rearrange the terms to get
$$
lambda(f(x_0)-r) > 0
$$
which implies that $lambda<0$. We may thus rescale $(x^*,lambda)$ so that $lambda=-1$.

Since $text{epi} f subset overline{text{epi} f} =overline{(text{epi} f^circ)} $ and that $(y,f(y))in text{epi}f$,
$$
langle x_0,x^* rangle - f(x_0) ge langle y,x^* rangle - f(y)
$$
for all $yin X$ (we take limit $rto f(y)$ so the strict $>$ becomes $ge$). Hence the (rescaled) functional $x^*$ is a subgradient of $f$ at $x_0$, i.e. $x^*in partial f(x_0)$.

In general, the statement is not true.

Consider a discontinuous linear functional $Lambda:Xto Bbb R$ (which can be constructed from a Hamel basis), since it is a linear functional it is automatically convex. However, at any point $xin X$ its subdifferential $partialLambda(x)$, i.e. the set of all $x^*in X^*$ such that
$$langle y - x, x^* rangle+ f(x) leq f(y) forall y in X,
$$
is empty.