Showing that $(\frac{\mathrm d}{\mathrm dx}+1)^{-1}$ is bounded on $L^2(0,\infty)$

Question

I wish to show that if we consider the symmetric operator $L= i\frac{\mathrm d}{\mathrm dx}$ on $L^2((0,\infty))$, then the resolvent $R(-i) = (L+i)^{-1}$ is everywhere defined on $L^2(0,\infty)$. This reduces to showing that for any $v\in L^2(0,\infty)$, we can solve for $u\in L^2(0,\infty)$ such that $$ \frac{\mathrm d u}{\mathrm dx}(x)+u(x)=v(x). $$ It is easy to see that we can use Duhamel's formula to get the formula $$ u(x)=\int_0^x\! e^{y-x}v(y)\,\mathrm dy. $$ so I wish to show that $T:v\mapsto u$ is bounded on $L^2$. I would like to apply Hilbert-Schmidt type bounds here, but unfortunately, the kernel $K(x,y)=1_{0\le y\le x}e^{y-x}$ does not lie in $L^2((0,\infty)^2)$. How would one prove that $T$ is bounded?

Answer 1

$$ |u(x)| \le\int_{0}^{x}e^{y-x}|v(y)|dy =\int_{0}^{x}e^{(y-x)/2}e^{(y-x)/2}|v(y)|dy $$ By Cauchy-Schwarz \begin{align} |u(x)|^2 & \le \int_{0}^{x}e^{y-x}dy\int_{0}^{x}e^{y-x}|v(y)|^2dy \\ & \le \int_{0}^{x}e^{y-x}|v(y)|^2dy \\ \|u\|^2 &\le \int_{0}^{\infty}\int_{0}^{x}e^{y-x}|v(y)|^2dy dx \\ &=\int_{0}^{\infty}\int_{y}^{\infty}e^{y-x}|v(y)|^2dxdy \\ &=\int_{0}^{\infty}\int_{y}^{\infty}e^{y-x}dx |v(y)|^2dy \\ &=\int_{0}^{\infty}|v(y)|^2dy \\ &=\|v\|^2. \end{align}