A question about continuous and convex functions

Question

Let $ f:[0,1]\rightarrow R $ a continuous function. It is true that, if $ f $ is also convex, than $$ f\left(\frac{a +b}{2}\right) \leq \frac{1}{b-a} \int_{a}^{b} f.$$ How to prove this? It holds a reverse of this statement, i.e. f is convex if $$ f\left(\frac{a +b}{2}\right) \leq \frac{1}{b-a} \int_{a}^{b} f ?$$ Thank you for the help.

Answer 1

Since ${1 \over 2}(a+b) = {1 \over b-a} \int_a^b x dx $, Jensen gives $f({1 \over 2}(a+b)) \le {1 \over b-a} \int_a^b f(x) dx $.

If you take $f_n$ to be the function defined by the graph (with $m={1 \over 2} (a+b)$): $(a,1), (m-{1 \over n}, 1), (m,0), (m+{1 \over n},1), (b,1)$ then it should be clear that $f_n$ is not convex and $f_n(m) = 0$ for all $m$ and $\int f_n >0 $.