Power Function is Convex Real Function
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Theorem
Let $p \ge 1$ be a real number.
Define $f : \hointr 0 \infty \to \hointr 0 \infty$ by:
- $\map f x = x^p$
for each $x \in \hointr 0 \infty$.
Then $f$ is a convex function.
Proof
Applying Derivative of Power twice, we have that:
- $f$ is twice differentiable on $\openint 0 \infty$
with:
- $\map {f' '} x = p \paren {p - 1} x^{p - 2}$
for each $x \in \openint 0 \infty$.
Since $p \ge 1$, we have:
- $p \paren {p - 1} \ge 0$
and so:
- $\map {f' '} x \ge 0$
for each $x \in \openint 0 \infty$.
From Real Function with Positive Derivative is Increasing:
- $f'$ is increasing on $\openint 0 \infty$
and so from Real Function is Convex iff Derivative is Increasing:
- $f$ is convex on $\openint 0 \infty$.
It remains to assure ourselves that:
- $\map f {t x + \paren {1 - t} 0} \le t \map f x + \paren {1 - t} \map f 0 = t \map f x$ for $x \in \hointr 0 \infty$ and $t \in \closedint 0 1$.
We have:
- $\map f {t x} = \paren {t x}^p = t^p x^p$ for $x \in \hointr 0 \infty$ and $t \in \closedint 0 1$.
Since $p \ge 1$, we have:
- $t^p \le t$ for $t \in \closedint 0 1$.
Hence we have:
- $t^p x^p \le t x^p$
Hence:
- $\map f {t x} \le t \map f x$
Hence:
- $f$ is convex.
$\blacksquare$