Prove that $(b - a)^p \leq b^p - a^p$ when $0 \leq a \leq b$ and $p \geq 1$

Question

I'm trying to prove the inequality $(b - a)^p \leq b^p - a^p$ when $0 \leq a \leq b$ and $p \geq 1$ (where $p$ is not necessarily an integer). One way to show this is via the computation $$ (b - a)^p = \int_0^{b - a} px^{p - 1}dx \leq \int_a^b px^{p - 1}dx = b^p - a^p, $$ where we use the fact that the derivative $px^{p - 1}$ is an increasing function. The above proof seems to rely on the convexity of $x \mapsto x^p$ for $p \geq 1$ because the derivative of a function is increasing if and only if it is convex and differentiable. In fact, the desired inequality fails when $p < 1$, where the map $x \mapsto x^p$ is no longer convex.

Is there a way to prove the above inequality that more directly utilizes the convexity of $x \mapsto x^p$? For instance, is there a way to prove the analogous inequality $$ f(b - a) \leq f(b) - f(a) $$ for a convex function $f : \mathbb{R} \to \mathbb{R}$, where $f$ is not necessarily differentiable? Also, are there any further generalizations of this inequality, given that the condition $0 \leq a \leq b$ is slightly awkward?

Answer 1

Going to answer my own question following @dxiv's comment. Given a function $f : [0, \infty) \to \mathbb{R}$, if we prove that $f$ is superadditive, i.e. $$ \tag{$\ast$} f(x + y) \geq f(x) + f(y) \quad \text{for all $x, y \in \mathbb{R}$,} $$ then we will have proved the inequality $ f(b - a) \leq f(b) - f(a) $ from my question.

To get superadditivity, we need only require that $f$ is convex and $f(0) \leq 0$. By definition of convexity, we have the inequality $$ f(tx) = f(tx + (1 - t) \cdot 0) \leq tf(x) + (1 - t)f(0) \leq tf(x) $$ for all $t \in [0, 1]$ and $x \in [0, \infty)$. For $x, y \in [0, \infty)$, we can take $t = \frac{x}{x + y}$ (or $t = 1$ if $x = y = 0$) to get $$ \begin{aligned} f(x) &= f(t(x + y)) \leq tf(x + y), \\ f(y) &= f((1 - t)(x + y)) \leq (1 - t)f(x + y). \end{aligned} $$ Adding these inequalities together, we get ($\ast$). Note that it is important the domain of $f$ is in $[0, \infty)$, else $t$ may not lie in the interval $[0, 1]$.

Answer 2

By letting $c=b-a$ this reduces to $\sqrt[p]{a^p + c^p} \le c + a$, or $\|(a,c)\|_p \le \|(a,c)\|_1$.

Generally, for $x \in \mathbb{C}^n$, we have $\|x\|_q \le \|x\|_p$, where $1 \le p \le q \le \infty$. It is straightforward to show that $|x_k| \le \|x\|_p$ for any $p \ge 1$ and so $\|x\|_\infty \le \|x\|_p$ follows, so suppose $q < \infty$.

Since norms are positive homogeneous, it is sufficient to show that $\|x\|_q \le 1$ for $\|x\|_p=1$.

Suppose $\|x\|_p = 1$. Note that for $t \in [0,1]$ we have $t^p \ge t^q$ and so $|x_k|^q \le |x_k|^p$ and so $\|x\|_q^q = |x_1|^q+\cdots+|x_n|^q \le |x_1|^p+\cdots+|x_n|^p = 1$ and so $\|x\|_q \le 1$.