Define $T:L^2(\mathbb R) \rightarrow L^2(\mathbb R)$ by $$(Tf)(x)=\int_{-\infty}^x e^{-(x-y)} f(y) \, dy.$$
I would like to show that $T$ is bounded and that $$\lambda = \frac{1}{1+iw}$$ is in its continuous spectrum for $w\in\mathbb R$.
I am stuck on the first part. It is not a Hilbert-Schmidt integral operator (the kernel is not in $L^2$), and Minkowski's integral inequality does not work, so I'm not sure how to proceed.
On
Your operator is $$Tf(x) = \int^\infty_{-\infty}1_{y\leq x}e^{-(x-y)}f(y)\,dy $$ which by change of variables $z=x-y$ becomes $$Tf(x) = \int^\infty_{-\infty}1_{z\geq 0}e^{-z} f(x-z)\,dz. $$ This is just the convolution operator $$f\mapsto \phi*f $$ with $$\phi(z) = 1_{z\geq 0}e^{-z}.$$
Now one of the fundamental inequalities (not hard to prove) about convolutions is $$\|\phi*f\|_2\leq \|\phi\|_1\|f\|_2.$$ Since $\phi$ is integrable on $\mathbb{R}$, this gives you the boundedness of the operator with norm at most $\|\phi\|_1$.
$Tf=\phi*f$, where $\phi(t)=e^{-t}\chi_{(0,\infty)}(t)$. (I had $\phi$ backwards in the first version; noticed that guest's $\phi$ was different, then noticed his was right.) So $$\widehat{Tf}=\hat\phi\hat f.$$You can easily calculate $\hat\phi$; now the norm of $T$ is $||\hat\phi||_\infty$ and the spectrum of $T$ is the essential range of $\hat\phi$. (In other words, the supremum and the range, since $\hat\phi$ is continuous.)
(Since $T$ is convolution with an $L^1$ function one can use Minkowski to show it's bounded. Rewrite the definition so it looks like an average of translates of $f$...)