# real analysis – Excercise about convex functions counterexample

I’m self-studying real analysis using A course in mathematical analysis by Garling

One of the problems after the subsection on convex functions defines a function as midpoint convex if $$f(frac{a+b}{2}) leq frac{f(a)+f(b)}{2}$$
The problem is then to prove that, if f is midpoint convex, $$hgt 0$$ and $$n in Bbb{N}$$frac{f(c-h)-f(c)}{n+1} leq f(c+frac{h}{n}) – f(c)\$\$
I don’t think this statement is true. Let \$f(x) = x^2,\$ \$n = 1,\$ \$ c = -1\$ and \$h = 0.25\$

f(x) is midpoint convex as $$frac{a^2+2ab+b^2}{4} leq frac{a^2+b^2}{2}$$ is equivalent to
$$2ab leq a^2+b^2$$ which is true but $$frac{f(-1.25)-f(-1)}{2} = 0.28125 gt -0.4375 = f(-0.75) – f(1)$$