How to prove $|a+b|^p\leq|a|^p+|b|^p$ for $0\leq p\leq 1$ from concavity of $x\mapsto x^p$ for $x\geq 0$

Question

In this answer, it is shown by elementary means that $|a+b|^p\leq|a|^p+|b|^p$ for $0\leq p\leq 1$. Can we show the same elementary fact by using concavity of $\phi(x)=x^p$ for $0\leq p\leq 1$, $x\geq 0$, either by using one of the equivalent definitions of concavity/convexity (including the one with the differential quotient and its monotonicity), the (reverse) Minkowski inequality or the (reverse) Jensen inequality? You can also use the multi-dimensional (reverse) Jensen inequality, for example with the function $\varphi(x,y)=(x^{1/p}+y^{1/p})^p$ for $x\geq 0$, $y\geq 0$, which is concave for $p\geq 1$ and convex for $0\leq p\leq 1$, but at least, I want to see a different proof of this fact which uses in some sense concavity/convexity. I find it intuitive to argue like this but I cannot make the argument rigorous.

Answer 1

Hint: concavity of $f(x)=x^p$ on $[0,1]$ gives for any $x,y>0$ $$\left(\frac{x}{x+y}\right)^p=f\left(\frac{x}{x+y}\right) \ge \frac{y}{x+y}f(0)+\frac{x}{x+y}f(1)=\frac{x}{x+y}$$ and similarly $\displaystyle \left(\frac{y}{x+y}\right)^p \ge \frac{y}{x+y}$.

Answer 2

First, recall that if $f$ is concave, then for $x < y < z$ in the domain of $f$, we have $\frac{f(y) - f(x)}{y - x} \ge \frac{f(z) - f(y)}{z - y}$. To see this, note that there is $t \in (0, 1)$ such that $y = (1-t)x + tz$. If we substitute, then we see that the desired inequality is equivalent to the definition of concavity, $f((1-t)x + tz) \ge (1-t) f(x) + t f(z)$.

You can use that fact to prove in turn that if $f : [0, \infty) \to \mathbb{R}$ is concave and $f(0) = 0$, then $f$ is subadditive, i.e. $f(a+b) \le f(a) + f(b)$ for $a, b \ge 0$. For an outline of the proof: by symmetry, we may assume $a \le b$. Also, the cases $a = 0$ and $b = 0$ are easy, so we might as well assume $a > 0$ and $b > 0$. Now, if $a = b$, then the desired inequality is a direct consequence of the fact that $\frac{f(0) + f(2a)}{2} \le f\left( \frac{0+2a}{2} \right)$. Otherwise, we have $$ \frac{f(a) - f(0)}{a - 0} \ge \frac{f(b) - f(a)}{b - a} \ge \frac{f(a+b) - f(b)}{(a+b) - b}. $$ Therefore, $f(a) - f(0) \ge f(a+b) - f(b)$, which easily implies $f(a+b) \le f(a) + f(b)$.

Alternately, if you know $f''(x) \le 0$, then if $a\le b$, you can apply the mean value theorem on the intervals $[0, a]$ and $[b, a+b]$, and then apply the fact that $f'$ is decreasing so $f'(\xi_1) \ge f'(\xi_2)$, to get the desired inequality.

Now, the original inequality from the post follows easily since $|a+b|^p \le (|a|+|b|)^p \le |a|^p + |b|^p$, where the second inequality comes from the above inequality for $f(x) = x^p$.

(Incidentally, the above fact is also useful in real analysis or topology for proving that if $d$ is a metric on some set, then both $d'_p(x,y) := (d(x,y))^p$ ($0 < p < 1$) and $d''(x,y) := \frac{d(x,y)}{1+d(x,y)}$ are also metrics on the same set. The latter is commonly used in cases where you want the metric to be bounded, e.g. in the proof that a countable product of metrizable spaces is metrizable.)

Answer 3

If $f:[0,\infty)\to\Bbb R$ is concave and $f(0)=0$ then $f$ is subadditive: $f(x+y)\le f(x)+f(y)$.

For concavity implies (indeed, is equivalent to) the fact that the secant line slope $$ {f(t)-f(s)\over t-s},\qquad s<t $$ is monotone decreasing simultaneously in $s$ and $t$. That is, increase either (or both) of $s$, $t$, and the slope above decreases. Use this with $(s,t) = (0,y)$ as compared to $(s,t) = (x,x+y)$, with $x>0, y>0$.

Alternatively, concave $f$ as above admits a right-hand derivative $f'_+(t)$ at each $t\ge 0$, such that $t\mapsto f'_+(t)$ is non-increasing and $f(b)-f(a) =\int_a^b f'_+(t)\,dt$ for $0\le a<b$. Then $$ f(x+y)-f(x) =\int_x^{x+y} f'_+(t)\,dt =\int_0^y f'_+(s+x)\,ds \le\int_0^y f'_+(s)\,ds =f(y)-f(0)=f(y). $$