Is the operator $T$: $Tf=\int_{0}^{\infty}f(x)\cdot\frac{y}{x^2+y^2}dx$ injective? Let the operator $T: D \to C(\mathbb{R})$, defined by $f \mapsto F(y)=\int_{0}^{\infty}f(x)\cdot\frac{y}{x^2+y^2}dx$. Is $T$ injective?
Here $D=\{f\in C(\mathbb{R}): |f(x)|\le1\; \forall x \in \mathbb{R}, \text{$f(x)$ is independent of $y$ and $y>0$}\}$.
 A: I can provide a partial answer that if $T$ is not injective, the function in the kernel must have unbounded support.
If $f$ is compactly supported on the interval $[a,b] \subset [0,\infty)$, then we can make the observation that the taylor series of the function $x \mapsto \frac{1}{y}\frac{1}{(x/y)^2 + 1}$ converges uniformly for $x \in [a,b], y \in [b,\infty]$ by the Weierstrass M-test. This implies that for $y > 2b,$
$$\int_a^b f(x)\frac{y}{x^2 + y^2}dx =
\sum_{n=0}^\infty \frac{1}{y^{2n+1}} \int_a^b f(x)x^{2n}dx.$$
This is a power series in $1/y$ that clearly converges for $y > 2b$. We also know that this series is 0 for all $y > 0$, so we conclude that this is a power series in $1/y$ that is 0 for $y > 2b$. By uniqueness of power series in an interval, this implies that the coefficients of said power series are 0.
We only need to check if the span of the even monomials is complete in $L^2([a,b])$. The monomials of even degree are clearly linearly independent, so we can extend them to an orthonormal basis by Gram-Schmidt. Let's apply Stone-Weierstrass: it's clear the set of monomials of even degree separate points on $[a,b]$. Thus, we see that this family is dense in $C([a,b])$.
At this juncture, we are done since we pick a sequence of polynomials $P_n \to f$ in the supremum norm, and note $$\|f - P_n\|_{L^2([a,b])} \leq C\|f - P_n\|_\infty \to 0.$$
We have thus shown the claim that $T$ is injective on compactly supported continuous functions.
In the above, we basically abused that we could consider $y > x$ in the integral, so that we could approximate $\frac{y}{x^2 +y^2}$ as a series and use power series shenanigans to get some orthogonality relations on a compact interval. I don't think any of these arguments could extend to the unbounded support case. At the very least, checking what happens if $f$ is Schwartz might be a place to start.
A: I can prove the claim for $f\in L^p(0,\infty)$ for any $1\le p<\infty.$ In particular the case of bounded functions with bounded supports is covered.
Assume  $F(y)=0$ for any $y>0.$ Then ${\rm Re}F(y)={\rm Im}F(y)=0$ for $y>0.$ Therefore it suffices to consider real valued functions $f.$ Let $g$ denote the extension of $f$ to the even function on $\mathbb{R}.$ Then $g\in L^2(\mathbb{R})$ and $$F(y)={1\over 2}\int\limits_{\mathbb{R}}g(x) {y\over x^2+y^2}\,dx$$ By the Plancherel identity we get (see)
$$0=\int\limits_{\mathbb{R}}\hat{g}(\xi)e^{-2\pi y|\xi|}\,d\xi\qquad y>0$$  We have  $\hat{g}\in L^q(\mathbb{R}),$ where $q=p/(p-1)$ for $1<p<\infty $ and $g\in C_0(\mathbb{R})$ for $p=1,$ where $C_0(\mathbb{R})$ denotes  the continuous functions vanishing at $\pm \infty.$ As $g$ is a real function its Fourier transform is even. Hence $$\int\limits_0^\infty\hat{g}(\xi)e^{-2\pi y\xi}\,d\xi=0,\qquad y>0$$ Thus the Laplace transform of $\hat{g}$ vanishes, which implies $\hat{g}=0$ and consequently $g=0.$
It is possible to apply  also the Stone-Weierstrass theorem at the last step. The linear span of the functions $y\mapsto e^{-2\pi y\xi},$ for $y>0,$ form an algebra separating points and nonvanishing in $C_0[0,\infty),$ the continuous functions vanishing at infinity. Therefore this algebra is dense in $C_0[0,\infty),$ Therefore it  is dense in $L^2(0,\infty).$ Thus $\hat{g}=0.$
