supremum of a function in a normed space

Question

Let $X$ be a normed space and $A \subset X$. Prove that: $$\sup(f(A))=\sup(f(cl(A))=\sup(f(conv(A)),$$ where $f \in X^{*}$. For the first equality, I thought to prove it by double inequality: Let $a=\sup(f(A))$, $b=\sup(f(cl(A))$ and $c=\sup(f(conv(A))$; we have that $A\subset cl(A)$, so $f(A)\subset f(cl(A))$, so $a \le b$; now, I don't know how to show that $a \ge b$; please, can somebody give me a suggestion?

Answer 1

Recall that $x\in\operatorname{cl}(A)$ means that every neighborhood $U\ni x$ contains a point from $A$.

So let $a=\sup f(A)$ and $b=\sup f(\operatorname{cl}(A))$. For any $\varepsilon>0$ there exists $x\in\operatorname{cl}(A)$ such that $$f(x)>b-\frac\varepsilon2.$$

The you can use a continuity of $f$ to find a neighborhood $U$ of the point $x$ such that $f(u)>f(x)-\varepsilon/2$ for any $u\in U$.

Now there exists a point $y\in A\cap U$ for which we have $$f(y) > f(x)-\frac\varepsilon2 > b-\varepsilon.$$

Therefore $a=\sup f(A) \ge b-\varepsilon$. Since this is true for every $\varepsilon>0$, you get $$a\ge b.$$

Notice that we only used continuity of $f$ here. When you try to prove the part about convex hull, you will use that $f$ is linear.

EDIT: You may notice that using very similar arguments we can show that $f(\operatorname{cl} A) \subseteq \operatorname{cl} f(A)$. It is well-known that this is equivalent characterization of continuity. See, for example, Prove that the following statements are equivalent characterizations of continuity and $f$ is continuous at $a$ iff for each subset $A$ of $X$ with $a\in \bar A$, $f(a)\in \overline{ f(A)}$..

An alternative way to show $\sup f(A) = \sup \operatorname{cl} f(A)$ could be based on this inclusion. Namely from $f(\operatorname{cl} A) \subseteq \operatorname{cl} f(A)$ you get $\sup f(\operatorname{cl} A) \le \sup (\operatorname{cl} f(A))$. It should be relatively easy to show that $\sup \operatorname{cl} B = \sup B$ holds for any $B\subseteq\mathbb R$. So you have $\sup(\operatorname{cl} f(A))= \sup f(A)$ and together you get $$\sup f(A) \le \sup f(\operatorname{cl} A) \le \sup (\operatorname{cl} f(A)) = \sup f(A).$$