$L^2$ mapping is necessarily onto or not?

Question

For $f \in L^2(\mathbb{R})$, let $$Tf(x) := \int_0^1 f(x+y)\,dy.$$Do we necessarily have that$$S: L^2(\mathbb{R}) \to L^2(\mathbb{R}),\text{ }Sf = f - Tf$$is onto?

Answer 1

This map is not onto.

The image consists of mean zero functions in the sense that $\int_{\mathbb{R}} Sf(x) dx=0$. Roughly speaking this is because

$$\int_\mathbb{R} Sf(x)dx=\int_\mathbb{R}\int_0^1 (f(x)-f(x+y)) dy dx = \int_0^1 \left( \int_\mathbb{R} f(x) dx - \underbrace{\int_\mathbb{R} f(x+y) dx}_{=\int_\mathbb{R} f(x) dx} \right) dy = 0. $$

Of course I cheated because the integral $\int_\mathbb{R} f(x) dx$ does not necessarily exist for a general $L^2$ function $f$.

However, you can make this argument precise by considering truncated integrals $\int_{-N}^M Sf(x) dx$ and using the Cauchy-Schwarz inequality to show that the integral $\int_{\mathbb{R}} Sf(x) dx$ does exist and is equal to zero.