Definition [Concave]: In a topological vector space $(V, \tau)$ ($\tau$ is neither discrete nor indiscrete), let $C$ be a convex subset and $f$ a real-valued function defined on $C$. $f$ is concave iff $$f(\lambda\,x_1 + (1 - \lambda)\,x_2) \geq \lambda\,f(x_1) + (1 - \lambda)\,f(x_2)\hspace{0.2in}\forall\,x_1, x_2\in C\hspace{0.1in}\text{and}\hspace{0.1in}\forall\,\lambda\in[0, 1]$$
This question is inspired by the proof of Theorem 1.24 included in this note. The proof assumed that in a locally convex topological vector space $(X, \tau)$, if $f$ is concave, then when $K$ is compact convex, $f(K)$ is bounded.
I would like to see a counter example if this is false but when I am trying to prove it correct, I can not proceed at some place and below is my attempt:
Proof. Assume by contradiction, there exists $x \in K$ such that $\sup_{y \in K}\,f(y) = \infty$. Then find $M \in \mathbb{N}$ such that $f(y_m) > M$. Pick $\lambda \in (0, 1)$, by concavity, we have:
$$f[\lambda\,x + (1 - \lambda)\,y_m] \geq \lambda\,f(x) + (1 - \lambda)\,f(y_m)$$
$$f[(1 - \lambda)\,x + \lambda\,y_m] \geq (1 - \lambda)\,f(x) + \lambda\,f(y_m)$$
Combine these two inequalities and then we have: $$ \begin{aligned} \begin{split} & \lambda\,f[\lambda\,x + (1 - \lambda)\,y_m] + (1 - \lambda)\,f[(1 - \lambda)x + \lambda\,y_m]\\ &\geq \lambda^2\,f(x) + \lambda\,(1 - \lambda)f(y_m) + (1 - \lambda)^2\,f(x) + \lambda\,(1 - \lambda)f(y_m)\\ &> \lambda^2\,f(x) + (1 - \lambda)^2\,f(x) + 2\,\lambda\,(1 - \lambda)\,M \end{split} \end{aligned} $$
Meanwhile, we have:
$$ \begin{aligned} \begin{split} &\lambda\,f[\lambda\,x + (1 - \lambda)\,y_m] + (1 - \lambda)\,f[(1 - \lambda)x + \lambda\,y_m]\\ &\leq f[\lambda^2\,x + (1 - \lambda)^2\,x + 2\,\lambda\,(1 - \lambda)y_m]\\ \end{split} \end{aligned} $$
Hence we have: $$ \begin{aligned} \begin{split} &\lambda^2\,f(x) + (1 - \lambda)^2\,f(x) + 2\,\lambda\,(1 - \lambda)\,M\\ & < f[\lambda^2\,x + (1 - \lambda)^2\,x + 2\,\lambda\,(1 - \lambda)\,y_m]\\ \end{split} \end{aligned} $$
Now let $\lambda = \frac{1}{n}\,(n \in \mathbb{N})$, then:
$$ \begin{aligned} \begin{split} &\frac{1}{n^2}\,f(x) + (1 - \frac{1}{n})^2\,f(x) + 2\,(1 - \frac{1}{n})\,\frac{M}{n}\\ & < f[\frac{1}{n^2}\,x + (1 - \frac{1}{n})^2\,x + 2\,(1 - \frac{1}{n})\,\frac{1}{n}\,y_m]\\ &\text{and we can see this holds for all $n \in \mathbb{N}$}\\ \end{split} \end{aligned} $$
Then I can not proceed because there is no guarantee the right hand side of the inequality will converge to $f(x)$.
This is not true. Take a vector space $X\ne\{0\}$ supplied with the indiscrete topology $\mathcal T = \{\emptyset, X\}$. Let $f\ne 0$ be a linear functional on $X$. Then $X$ is compact and convex, but $f(X) = \mathbb R$.
If you are willing to assume that the space is Hausdorff and $f$ is upper semicontinous (i.e. $\{x:\ f(x)\ge c\}$ are closed for all $c$) then the claim is true. However, is has nothing to do with concavity/convexity.