Continuity and convexity of $F(x)=\frac{1}{x}\int_{0}^xf(t)dt$

Question

Let $f:(0,+\infty)\rightarrow\mathbb{R}$ be continuous and convex. Show that $$F:(0,+\infty)\rightarrow\mathbb{R}$$$$F(x)=\frac{1}{x}\int_{0}^xf(t)dt$$ is continuous and convex.

My attempt: Continuity it's straight forward, since $f$ it's continuous it's also integrable which implies that $\int_{0}^xf(t)dt$ it's continuous over $(0,+\infty)$ and therefore $F$ it's the product of two continuous functions (since $1/x$ it's continuous over that same interval) hence $F$ itself it's continuous.

When it comes to convexity I tried calculating the second derivative of $F$, I got: $$F'(x)=\frac{1}{x}\left(f(x)-\frac{1}{x}\int_{0}^xf(t)dt\right)$$ $$F''(x)=\frac{1}{x}\left(\frac{2}{x^2}\int_{0}^xf(t)dt-\frac{2}{x}f(x)+f'(x)\right) $$

Since $\frac{1}{x}$ it's positive on the desired interval I tried analyzing the expression inside the parenthesis but without much success (I noticed I haven't used anywhere that $f$ is convex so that must be the key but I'm not sure how to utilize it). I also tried proving the inequality $F(\lambda x+(1-\lambda) y)\leq \lambda F(x) + (1-\lambda)F(y) $ but after substituting $F$ with it's definition the result looks messy and complicated... Any help?

Answer 1

Make the change of variable $t=xy$. Then your integral becomes $$ \int\limits_0^1 f(xy)dy $$ And its second derivative in $x$ Is $$ \int\limits_0^1 y^2f’’(xy)\,dy $$ Which is nonnegative.

Answer 2

Convex functions need not be twice differentiable so your approach is not valid. $F(x)=\int_0^{1} f(xy)dy$ by the substitution $y=\frac t x$. This makes it obvious (from definition) that $F$ is convex: $$F(tu+(1-t)v)=\int_0^{1} f((tu+(1-t)v)y)dy$$ $$ \leq \int_0^{1} [tf(uy)+(1-t)f(vy) dy =tF(u)+(1-t)F(v).$$