Let $m : \mathbb{R} \to \mathbb{R}$ be a continuous function of compact support. Given $M$ as:
$$ M(x) = \int_{\mathbb{R}} m(t) (t+x)^2\ \mathrm{d}t,$$
is it possible recover $m$?
I thought it should be possible via the convolution theorem, that is,
$$ M(x) = (m * g)(-x) = \int_{\mathbb{R}}m(t)g(-x-t)\ \mathrm{d}t, $$
where $g(-x-t) = (t+x)^2$, that is $g(y) = y^2$.
Let $\tilde{M}(x) = M(-x)$ and $\mathcal{F}$ denote the Fourier transform, then $\mathcal{F}(\tilde{M}) = \mathcal{F}(m)\cdot \mathcal{F}(g)$ and
$$m = \mathcal{F}^{-1}\left(\frac{\mathcal{F}(\tilde{M})}{\mathcal{F}(g)}\right).$$
The problem is that $\mathcal{F}(g)$ might be zero (actually I think it is Dirac-delta related). Is there any way to fix this, or maybe some other approaches to recovering $m$ from $M$? It does not have to be theoretical, any numerical solutions that allow for arbitrary precision are great as well.
Thanks in advance ;-)
The answer is no: we can find a continuous function with compact support such that $\int_{\mathbb R}f(x)\mathrm dx=0$, $\int_{\mathbb R}xf(x)\mathrm dx=0$ and $\int_{\mathbb R}x^2f(x)\mathrm dx=0$ with $f\neq 0$ identically.
Indeed, we can take an odd function (the first and third condition are satisfied) and $\int_{\Bbb R_+}xf(x)\mathrm dx=0$.