Surjectivity of an operator on $L^2(R)$ defined as identity minus convolution I'm having trouble with the following exercise. Define $T:L^2(\mathbb{R})\to L^2(\mathbb{R})$ by means of the formula
$$(Tf)(x)=\int_{0}^{1}f(x+y)dy.$$
I proved that, for all $f\in L^2(\mathbb{R})$, $Tf$ is a continuous function and $||Tf||_2\le||f||_2$, with equality iff $f(x)=0$ almost everywhere.
Then there is the following question, on which I'm stuck: is $S:=I-T$ surjective? Here, $I$ is the identity operator. This should be doable at the level of "Papa Rudin".
Edit:
Thanks to the last paragraph in @FTP's great answer, I think I have an idea for a more "elementary" (but more complicated) solution. 
We know that $X:=L^1(\mathbb{R})\cap L^2(\mathbb{R})$ is dense in $L^2(\mathbb{R})$. Suppose $S$ is surjective. Since $S$ is injective and continuous, by the Open Mapping Theorem, it follows that $S$ is invertible. 
Now, let $g\in X$ and $f=S^{-1}g$. By density, there is $\{f_n\}\subset X$ such that $f_n\to f$ in $L^2(\mathbb{R})$. Write $g_n=Sf_n$. Since each $f_n$ is integrable, it follows by Fubini-Tonelli that each $g_n$ is integrable, and we actually have
$$\int_{\mathbb{R}}g_n(x)dx\int_{\mathbb{R}}(Sf_n)(x)dx=0.\quad (*)$$
Also, by the continuity of $S$, we obtain that $g_n\to g$ in $L^2(\mathbb{R})$. 
If we choose $g$ very nice with positive integral, can we obtain contradiction with equation (*)?
 A: The operator $T$ is the convolution of $f$ with function $g=\chi_{[-1, 0]}$. Therefore, on the Fourier transform side we have $\widehat{Tf} = \hat f\hat g$. Here
$$
\hat g(\xi) = \int_{-1}^0 e^{-2\pi i \xi x}\,dx = \frac{e^{2\pi i \xi}-1}{2\pi \xi i}
\tag1 $$
Notably, $\hat g$ is smooth and $\hat g(0)=1$. We don't really need (1) to see this, the obvious properties that $g$ has compact support and $\int g=1$ are enough.
So, $I-T$ is $\hat f\mapsto (1-\hat g)\hat f$ on the Fourier side. This is clearly injective, as $1-\hat g \ne 0$ almost everywhere. If this map was  surjective, it'd be invertible (open mapping theorem), and the inverse can be nothing else but  $\hat f\mapsto (1-\hat g)^{-1} \hat f$. But   the latter is not a bounded operator on $L^2$  since $(1-\hat g)^{-1} \notin L^\infty$.
Without OMT
One can avoid the open mapping theorem and simply present a function $\phi\in L^2$ such that $\phi/(1-\hat g)\notin L^2$. For example, 
$$
\phi(\xi) = \max(0, |\xi|^{-1/3}-1)
$$ 
would do. Let $\check \phi$ be the inverse Fourier transform of $\phi$; then $\check \phi\notin \operatorname{ran}(I-T)$. Indeed, if $\check \phi = (I-T)f$, then $\hat f(\xi) = \phi(\xi)/(1-\hat g(\xi))$ which is not in $L^2$.
Without Fourier transform
Informally, the above is a long-winded justification of "since $Tf$ and $f$ have the same integral, the integral of $f-Tf$ must be zero". The preceding sentence doesn't make a proof because $f$ and $Tf$ need not be in $L^1(\mathbb{R})$. But one can consider the regularization 
$$
\lim_{c\to 0^+} \int_\mathbb{R} e^{-cx^2} (f(x)-Tf(x))\,dx
\tag 2$$
and show (with some help from Fubini) that (2) is $0$ for every $f\in L^2$. Then it becomes obvious that, for example, any positive function cannot be in $\operatorname{ran}(I- T)$. 
