I have some questions about the following two operators.
- A convolution operator $T$. If $k \in \mathcal L^1(\mathbb R)$, then $$f(x) \mapsto \int_{-\infty}^\infty k(x-y)f(y) dy: \mathcal L^2(\mathbb R) \to \mathcal L^2(\mathbb R).$$ How to see that this operator is well-defined? That is, why $Tf \in \mathcal L^2$? Moreover, how to see this operator is bounded, please?
- Integration on $\mathcal L^2[0,1]$: $$f(x) \mapsto \int_0^x f(s) ds.$$ Again, to see this operator is bounded, I have the following reasoning. $$\left| \int_0^x f(s) ds \right| \leq \int_0^x \left| f(s)\right| ds \leq \int_0^1 |f(s)| ds \leq \|f\|\cdot\|1\|.$$ Is my argument correct, please?
If $k \in L^{1}(\mathbb{R})$ and $f\in L^{2}(\mathbb{R})$, then $$ \left|\int_{-\infty}^{\infty}k(x-y)f(y)\,dy\right|^{2} \le \left(\int_{-\infty}^{\infty}|k(x-y)|^{1/2}|k(x-y)|^{1/2}|f(y)|\,dy\right)^{2} \\ \le \left( \int_{-\infty}^{\infty}|k(x-y)|\,dy\right) \left(\int_{-\infty}^{\infty}|k(x-y)||f(y)|^{2}\,dy\right) \\ \le \|k\|_{L^{1}}\int_{-\infty}^{\infty}|k(x-y)||f(y)|^{2}\,dy. $$ Now integrate both sides with respect to $x$ and use Fubini's Theorem to obtain $$ \|Tf\|_{L^{2}}^{2} \le \|k\|_{L^{1}}\int_{\infty}^{\infty}\left(\int_{-\infty}^{\infty}|k(x-y)|\,dx\right)|f(y)|^{2}\,dy \\ = \|k\|_{L^{1}}^{2}\int_{-\infty}^{\infty}|f(y)|^{2}\,dy=\|k\|_{L^{1}}^{2}\|f\|_{L^{2}}^{2}.\\ \implies \|Tf\|_{L^{2}} \le \|k\|_{L^{1}}\|f\|_{L^{2}}. $$ Your argument in part (2) is just fine, but, as a final step, you do have to integrate the square of $|\int_{0}^{x}f(s)\,ds|$ with respect to $x$.