Weird property of $f(x) + f(\sqrt{1-x^2})$ on the interval [0,1]

Solution 1:

If you make a change of variable $x \mapsto \cos \theta$ then you are looking at a function of the form $g(\theta) + g(\frac \pi 2 - \theta)$ on the interval $[0, \frac \pi2]$, where $g(\theta) = f(\cos(\theta))$.

This function is always symmetric with respect to $\theta = \frac\pi 4$. If you want $g$ to be differentiable at $\theta = \frac\pi 4$, then the derivative there is necessarily $0$.

Even though, it is not necessary that the function has even a local maximum/minimum at $\theta = \frac \pi 4$.

Example: take $g(\theta) = (\theta - \frac\pi4)^2\cos(1/(\theta - \frac\pi 4))$. You can verify that $g(\frac\pi2 - \theta) = g(\theta)$, so that $g(\theta) + g(\frac \pi 2 - \theta) = 2g(\theta)$.

Now here is a plot of $g$:

Solution 2:

Neat property! Here's why it works (for some of the time):

Take $f$ to be continuous. Set $g(x) = f(x) + f(\sqrt{1 - x^2})$.

If it so happens that $f$ is differentiable at $x$, then you could probably see (with a little bit of chain rule)

$$g'(x) = f'(x) - \frac{xf'(\sqrt{1-x^2})}{\sqrt{1-x^2}}.$$

Now, if $f$ is differentiable at $x = 1/\sqrt2$, then the above tells us that $g'(1/\sqrt2) = 0$. So, if $f$ is differentiable at $1/\sqrt2$, then the derivative of $g$ must be $0$ at $1/\sqrt2$.

From calculus, we'd interpret this by $g$ having a local max or min at $1/\sqrt2$, which you're able to guarantee if $f$ is differentiable at $1/\sqrt2$.

However, this property is not true necessarily for all continuous $f$: in particular those $f$ that are not differentiable at $1/\sqrt2$. Try it out with some functions like

$$f(x) = |x - 1/\sqrt2| \quad \text{or} \quad f(x) = \begin{cases}\left(x - \frac1{\sqrt2}\right)\sin\left(\frac1{x-1/\sqrt2}\right) & \ x \neq 1/\sqrt2 \\ 0 &x = 1/\sqrt2 \end{cases}$$

In particular, the second function above looks kinda crazy, but is in fact continuous on all of $\mathbb R$! See what happens when we graph $g$ for these $f$.

A similar property also holds for $h(x) = f(x)+f(-x)$, which can also be kinda neat.