Simple-looking problem with integrals

Question

Let $f: [0,\infty) \rightarrow \mathbb{R}$ be a continuous function such that $f(0) = 0$. Is it true that if the integral $$ \int_0^{\pi/2} \sin(\theta) f(\lambda \sin(\theta)) \, d\theta $$ is zero for every $\lambda > 0$, then $f$ is identically zero?

It's rather obviously true if $f$ is a polynomial and I'm hoping it is true in general, which is perhaps why I'm stuck.

Edit. I came across this problem in two different, but related contexts. I'll describe the easier one: given a positive continuous function $F : \mathbb{R} \rightarrow \mathbb{R}$, the functions $\kappa_\lambda(\theta) := F(\lambda\cos(\theta))$, with $\lambda > 0$, are all curvature functions of plane ovals evaluated at the point of the curve where $(\cos(\theta),\sin(\theta))$ is the normal vector if and only if $$ \int_0^{2\pi} e^{-i\theta} \kappa_\lambda(\theta) \, d\theta = \int_0^{2\pi} e^{-i\theta} F(\lambda\cos(\theta)) \, d\theta = 0 $$ for all $\lambda > 0$. If $F$ is even, this is always the case, but does it have to be even? Well, after you decompose $F$ into even and odd parts and play around with this you come to the problem posed above and so nicely solved by Fedor Petrov and Mateusz Kw'asnicki below. Their solution readily implies that

A continuous function $F : \mathbb{R} \rightarrow \mathbb{R}$ is even if and only if $$ \int_0^{2\pi} e^{-i\theta} F(\lambda\cos(\theta)) \, d\theta = 0 $$ for all $\lambda > 0$. In other words, all the ovals will be centrally symmetric.

Answer 1

As suggested by Fedor Petrov, we write $$ g(x) = f(\sqrt x) , $$ and we substitute $\lambda \sin\theta = \sqrt{x}$ and $t = \lambda^2$. This leads to $$ \begin{aligned} 0 & = 2 \lambda \int_0^{\pi/2} f(\lambda \sin \theta) \sin \theta \, d\theta \\ & = \int_0^{\lambda^2} 2 f(\sqrt x) \sqrt{x} \, \frac{1}{2 \sqrt{x (\lambda^2 - x)}} \, dx \\ & = \int_0^t \frac{g(x)}{\sqrt{t - x}} \, dx . \end{aligned} $$

Now, this is the "half-integral" of $g$: the convolution of $g$ with $x^{-1/2}$ is the fractional Riemann–Liouville integral of $g$, up to a constant factor. Adding another half-integral leads to the usual integral: by Fubini, $$ \begin{aligned} 0 & = \int_0^s \biggl(\int_0^t \frac{g(x)}{\sqrt{t - x}} dx\biggr) \frac{1}{\sqrt{s - t}} dt \\ & = \int_0^s g(x) \biggl(\int_x^s \frac{1}{\sqrt{t - x}} \frac{1}{\sqrt{s - t}} dt \biggr) dx . \end{aligned} $$ The inner integral is just $\pi$: substituting $t = x + (s - x) u$, we obtain $$ \int_x^s \frac{1}{\sqrt{t - x}} \frac{1}{\sqrt{s - t}} dt = \int_0^1 \frac{1}{(s - x) \sqrt{u (1 - u)}} \, (s - x) \, du = \pi .$$ It follows that $$ 0 = \pi \int_0^s g(x) dx , $$ which clearly implies that $g$ is zero almost everywhere. Edit: ...and since $g$ is continuous, almost everywhere yields everywhere (as pointed out by Fedor Petrov in a comment below).