Kernel of an integral operator with Gaussian kernel function Suppose we have the integral operator $T$ defined by
$$Tf(y) = \int_{-\infty}^{\infty} e^{-\frac{x^2}{2}}f(xy)\,dx,$$
where $f$ is assumed to be continuous and of polynomial growth at most (just to guarantee the integral is well-defined). If we are to inspect that the kernel of the operator, we would want to solve
$$0 = \int_{-\infty}^{\infty} e^{-\frac{x^2}{2}}f(xy)\,dx.$$
If $f$ were odd, this would be trivially zero so we would like to consider the case that $f$ is even. My hunch is that $f$ should be identically zero but I haven't been able to convincingly prove it to myself. The reason that I feel like it should be the zero function is that by scaling the Gaussian, we can make it arbitrarily close to $0$ or $1$. I'll sketch some thoughts of mine.
If $y=0$, then we would have that $0 = \sqrt{2\pi}f(0)$, forcing $f(0)=0$. Making use of even-ness and supposing instead that $y\neq 0$, we can make a change of variable to get
$$0 = \int_0^{\infty} e^{-\frac{x^2}{2y^2}}f(x)\,dx.$$
Since $f$ has polynomial growth at most, given a fixed $y$, for any $\varepsilon > 0$, there exists $R_y$ such that
$$\left|e^{-\frac{x^2}{2y^2}}f(x)\right| \le \frac{\varepsilon}{1+t^2}$$
for all $x > R_y$. Thus we can focus instead on the integral from $0$ to $R_y$ since the tail effectively integrates to zero:
$$0 = \int_0^{R_y}e^{-\frac{x^2}{2y^2}}f(x)\,dx.$$
Since $[0,R_y]$ is compact and $f$ is a continuous function, it can be approximated uniformly by (even) polynomials with constant term $0$ by Stone-Weierstrass, i.e.
$$f(x) = \lim_n p_n(x),$$
where $p_n(x) = \sum\limits_{m=1}^n a_{m,y}x^{2m}$. Here the coefficients are tacitly dependent upon the upper bound (so I've made it explicit to prevent any confusion). From here, we have
$$0 = \int_0^{R_y}e^{-\frac{x^2}{2y^2}}\lim_n p_n(x)\,dx.$$
However since the convergence is uniform, we can commute limit and integral to get that
$$0 = \lim_n\sum_{m=0}^n a_{m,y}\int_0^{R_y}e^{-\frac{x^2}{2y^2}}x^{2m}\,dx.$$
Making use of a change of variable, this gives
$$0 = \lim_n\sum_{m=0}^n a_{m,y}y^{2m+1}\int_0^{\frac{R_y}{y}} e^{-\frac{x^2}{2}}x^{2m}\,dx.$$
I would like to be able to say that the coefficients must be zero but this is pretty messy at this point. Does anyone have any clue as to how to proceed? Or is there a better way to do this? (I would like to avoid Fourier transform-based or Weierstrass transform-based arguments.)

easy there are infinity solutions like $$y^2 \sin (x)+x \cos (x)$$ $$e^{-x} y^2+e^{-x} x$$
done easy
