Consider the operator $A:\mathcal L_{2,w}(\mathbb R)\to\mathcal L_{2,w}(\mathbb R)$, which maps from the weighted $\mathcal L_2$-type space to itself. The operator acts in the following way: $$(Af)(t)=(\chi_{[0,\infty)}*f)(t)=\int_{-\infty}^tf(s)ds,$$ and can be thought of as a convolution operator where one of the convolved functions is fixed as the indicator function on $[0,\infty)\subset\mathbb R$. The norm on this space is an extension of the usual $\mathcal L_2$-norm and is given by $$\|f\|_{\mathcal L_{2,w}}=\left(\int_{\mathbb R}|f(t)|^2e^{-2wt}\text{d}t\right),$$ where $w\in\mathbb R_{\ge0}$ is a fixed parameter.

I want to find the operator norm of $A$. Using Young's inequality I have been able to show that $\|A\|\le\|\chi_{[0,\infty)}\|_{\mathcal L_{1,w}}=\frac{1}{|w|}$, but am not sure on showing the other direction. I have found this question here to be quite similar to mine, and am wondering if there is an obvious choice of $f\in\mathcal L_{2,w}$ to make use of here. On the other hand, I wonder if I should make use of an approximate identity or similar.

Any suggestions would be appreciated.

  • $\begingroup$ How is the $\mathcal L_1$-norm of $\chi_{[0,\infty)}$ is $1/|w|$? $\endgroup$ – Berci May 22 at 11:20
  • 2
    $\begingroup$ Remember that this is the weighted $\mathcal L_{1,w}$-type norm. We have: $\|\chi_{[0,\infty)}\|_{\mathcal L_{1,w}}=\int_\mathbb R|\chi_{[0,\infty)}(t)|e^{-wt}dt=\int_0^\infty e^{-wt}dt=\left[\frac{-1}{|w|}e^{-wt}\right]_0^\infty=\frac{1}{|w|}$. $\endgroup$ – Sarah Paulson's Scream May 22 at 11:44
  • $\begingroup$ Ahh, ok.. I somehow misread it and thought there's also an $i$ in the exponent. $\endgroup$ – Berci May 22 at 11:50

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