Prove that midpoint convex and bounded imply continuous

Question

Question: Let $f$ be a function on $[a, b]$. For $\forall x\in[a, b], |f(x)|\le M$ where $M>0$, and for $\forall x, y\in[a, b], f(\frac{x+y}{2})\le\frac{f(x)+f(y)}{2}$.

(1) Prove that $f$ is continuous on $(a, b)$.

(2) Must $f$ be continuous at $a$ and $b$, respectively?

Note that I already checked the answer in:

Does $f(\frac{x+y}{2})\leqslant \frac{f(x)+f(y)}{2}$ imply continuity?

Example of a function such that $\varphi\left(\frac{x+y}{2}\right)\leq \frac{\varphi(x)+\varphi(y)}{2}$ but $\varphi$ is not convex

Midpoint-Convex and Continuous Implies Convex

but none of them helped. The answer in the first one is not a precise solution I think. He(or she) put $x_0+\delta (\delta>0)$ as $x_0+$ and regarded $x_0+\delta /2(\delta>0)$ as $x_0+$, which is non-sense. So, it was quite weird to me.

I knew that without the condition that $f$ is bounded, we cannot say that $f$ is convex (or continuous). Instead, when $f$ is continuous unless $f$ is bounded, then $f$ is convex. (This is actually not a key point of this problem, I think; just interesting stuff)

The closest solution was in:

Proof every convex function is continuous (Problem 10 Convex Functions Spivak)

They took account of the general form of Jensen's Inequality, but in this problem, we only have to deal with $\lambda=1/2$. I tried in an analogous way, but I stuck:

For $\forall x, y\in [a, b]$, let $z$ be in $[a, b]$ such that $y=\frac{x+z}{2}$. Substituting $x=x, y=z$ will bring $$f(y)=f(\frac{x+z}{2})\le \frac{f(x)+f(z)}{2}= f(x)+\frac{f(z)-f(x)}{2}$$ $$f(y)-f(x)\le{1\over2}(f(z)-f(x))\le{1\over2}(M-(-M))\le M$$

I do want to make the form $f(y)-f(x)\le K|y-x|$, which implies $f$ is Lipschitz continuous(and of course, continuous on $[a, b]$; we can prove this by setting $\delta=\epsilon/K$. But I failed to make such a form.

Could you give me some hints or other ideas to the problem? Thanks.

(+) I think there are many counterexamples of (b).