Prove that $f:[0,\infty) \to \mathbb{R}, x\mapsto x^p$ for $p\geq 1$ is convex.

Question

Prove that $f:[0,\infty) \to \mathbb{R}, x\mapsto x^p$ for $p\geq 1$ is convex.

I tried using this chain of inequalities:$$\frac{f(x_2)-f(x_1)}{x_2-x_1}\leq \frac{f(x_4)-f(x_3)}{x_4-x_3}\leq \frac{f(x_6)-f(x_5)}{x_6-x_5}$$ if $a<x_1<x_2\leq x_3 <x_4 \leq x_5 <x_6<b$. Any Ideas?

Answer 1

HINT.-Applying a known theorem about convexity, the proof is immediate since $p\ge1$ (Take a look at Aqua's answer). If you want to explicit a proof for the particular case of your exponential functions you can do as follows (among other possibilities).

►It is easy to verify that the arc of the curve between the origin and an arbitrary point $(a, f (a))$ is convex because this segment has equation $y(x) =\dfrac {f (a)}{a}x$ for $0\le x\le a$.

►It follows that if $b\gt a$ the arc of the curve from $(a,f(a))$ to $(b,f(b))$ is also convex because if not then there is a point $c$ in the interval $[a,b]$ such that $f(c)$ two distinct values. Can you see why? Help you with the attached figure.