Let $C$ be a closed bounded convex subset of some Banach space $E$ (alternatively locally convex), let $f:E\to\mathbb R$ be a continuous affine functional, then is it true that there exist $a$ and $b$ in $\mathbb R$ such that $f(S)=[a,b]$ ?
Here is something I tried. If we have an infinite set $\{ e_n \}_{n\in\mathbb N}$ of independent unit vectors in some Hilbert space, suppose we take the closed unit ball, then it is closed bounded and convex. Now let $f(x)=\sum_{i\in\mathbb N} \langle x , e_i\rangle$, let $x_n=\frac{1}{\sqrt{n}}\sum_{k=1}^n e_k$, then $\|x_n\|=1$ for any $n$ and $f(x_n)=\sum_{i\in \mathbb N} \langle x_n , e_i\rangle=\sqrt{n}$, therefore $f(B_1) = ]-\infty,\infty[$.
Is this reasoning correct ?
Best Answer
This does not work without some kind of compactness.
Let $C$ be the closed unit ball of $C([0,1])$ and define $$ f(x):= \int_0^{1/2} x(t)dt - \int_{1/2}^1 x(t)dt. $$ Then $f(C) = (-1,+1)$.