Consider a linear program $$f(b)=\min_{x}\{c^\top x: A x = b, x\geq 0\}$$ (assume it is feasible and bounded for all $b$). My understanding is that $f(b)$ is a convex piecewise linear function of $b$ (can be shown taking the dual of the LP, see below).
My question
Can I also say that $f(b)$ is continuous? Or when can I say that?
Below my attempts to workout a proof, though I strongly believe someone has done that already.
Proof that $f(b)$ is convex and piecewise linear
By strong duality $$f(b)=\min_{x}\{c^\top x: A x = b, x\geq 0\}=\max_\pi\{\pi b:\pi A\leq c\}$$ If we let $\pi_1,\ldots,\pi_K$ be the extreme points of $\{\pi A\leq c\}$ we can rewrite the dual as $$f(b)=\max_{\pi_1,\ldots,\pi_K}\{\pi_i b\}$$ This gives us a piecewise linear function in $b$. Furthermore, the pointwise max of a set of linear functions is convex, hence $f(b)$ is convex and piecewise linear in $b$.
Working out a proof
Assume we take $b\in B$ where $B$, loosely speaking is the set of vectors $b$ for which the linear program has a finite optimal solution, i.e. $f(b)$ is defined.
Now, since linear functions are continuous, if the pointwise maximum of continuous functions is continuous the proof is done (but I do not know if that holds). I will try in another way:
Take a sequence $(b_n)$ converging to $b_0$ (assume $B$ is closed). I need to show that $\lim f(b_n)=f(b_0)$ $$\begin{align} \lim f(b_n) =& \lim\max_{\pi_1,\ldots,\pi_K}\{\pi_i b_n\}\\ =&\max_{\pi_1,\ldots,\pi_K}\{\lim\pi_i b_n\}\\ =&\max_{\pi_1,\ldots,\pi_K}\{\pi_i\lim b_n\}\\ =&\max_{\pi_1,\ldots,\pi_K}\{\pi_i b_0\}\\ =&f(b_0) \end{align}$$ Hence $f(b)$ is continuous. But I am walking on thin ice when I say that the limit of the maximum is equal to the maximum of the limit (second line).