For which values of $q\ge1$ is $x\mapsto |x-a|^q$ strictly convex?

Question

For which values of $q \ge 1$ is $f(x) = |x-a|^q$ strictly convex with $a \in \mathbb{R}$ fixed? I know that if $q>1$, we have:

$\frac{d^2}{dx^2} |x-a|^q = q(q-1)|x-a|^{q-2} > 0$ for all $x \ne a$

So $f$ is strictly convex on $(-\infty,a)$ and $(a,\infty)$. But what about at $x=a$? Also what if $q=1$?

Is the answer to the question simply that $f$ is strictly convex for $q>1$ and convex for $q=1$?

Thanks for any help you can give!

Answer 1

Without loss of generality (using a translation), we can suppose that $a=0$.

For $q=1$, $f_q(x)= \vert x \vert^q$ is not strictly convex as for $x_1 =0$, $x_2=x>0$ and $\lambda \in (0,1)$, we have $f_1((1-\lambda)x_1 +\lambda x_2)= \lambda x_2$ without a strict inequality.

Now for $q > 1$ as you noticed, $f_q$ is strictly convex on both $(-\infty,0]$ and $[0,\infty)$. So let’s take $x_1<0<x_2$ and $\lambda \in (0,1)$. If $(1-\lambda)x_1 +\lambda x_2 <0$ we have $$f_q((1-\lambda)x_1 +\lambda x_2) < (1-\lambda)f_q(x_1) < (1-\lambda)f_q(x_1) +\lambda f_q(x_2)$$

The first inequality follows from the strict convexity of $f_q$ on $(-\infty,0]$ and the second one as $f_q(x_2) >0$.

As we can deal in a similar way for $(1-\lambda)x_1 +\lambda x_2 >0$, this complete the proof that $f_q$ is strictly convex.

Answer 2

Spivak's book states first this definition of Convex Function:

Definition 1:

A function $f$ is convex on an interval, if for all $a$ and $b$ in the interval, the line segment joining $(a, f(a))$ and $(b, f(b))$ lies above the graph of $f$.

Now, this means that the stright line defined by the function $g(x)=\frac{f(b)-f(a)}{b-a}(x-a)+f(a)$ is such that $f(x)\leq(x)$. But this is equivalent to say that $\frac{f(x)-f(a)}{x-a}\leq\frac{f(b)-f(a)}{b-a}$. This is the reason why then it states an equivalent definition.

Definition 2:

A function $f$ is convex on an interval if for $a$, $x$, and $b$ in the interval with $a<x<b$ we have $\frac{f(x)-f(a)}{x-a}\leq\frac{f(b)-f(a)}{b-a}$.

From these definitions, it is not difficult to see that convexity is invariant to linear change of variable with positive Jacobian, i.e. positive slope, hence by $x\to x+a $ it is enough to show $f(x) = |x|^q$ is convex.

By Definition 2, this corresponds to the following inequality: $$\frac{|x|^q-|a|^q}{x-a}\leq\frac{|b|^q-|a|^q}{b-a}$$

Which is a case of Cauchy Schwarz inequality, that can easily be proven.

Another way of showing that is as follows: Referring to definition 1, for $q\geq 1$ and $a,b>$ of the same sign the statement in definition 1 is easy using the derivative method you used, and for case $a<0<b$ the argument can be split as follows:

$f$ is convex between on $(a,0)$ hence the line segment joining $(a, f(a))$ and $(0, f(0))$ lies above the graph of $f$. The same goes for $(0,b)$ and the line segment joining $(0, f(0))$ and $(b, f(b))$, but since $f(x)\geq f(0)=0$ for all $x\in \mathbb{R}$, the line segment joining $(a, f(a))$ and $(b, f(b))$ lies above both the previous line segments, hence our function is indeed convex.

Answer 3

Hint: you may start with the first derivative $$ \frac{d}{dx}|x-a|^q=q|x-a|^{q-1}\operatorname{sign}(x-a) $$ and show that for $q>1$ it is continuous and strictly increasing.