Task 1. Let $S, T: L^2(0, \infty) \to L^2(0, \infty)$ be given by $(Sf)(t) = f(2t), (Tf)(t) = f(t/2)$, $f$ is an element of $L^2(0, \infty)$. You are given that $S$ and $T$ are linear mappings $L^2(0, \infty) \to L^2(0, \infty)$, and you need not prove this fact.

(i) Calculate $\| S\|$.

(ii) Show that $T^{*} = 2S$. Determine $S^{*}$.

Solution. (i) Take an arbitrary $f \in L^{2}(0,\infty)$ and consider

Make the substitution $s = 2t$. Then $t = \frac{s}{2}$ and $dt = \frac{ds}{2}$. If $t$ changes from 0 to $\infty$, then $s$ also changes from 0 to $\infty$. This means that after the substitution we obtain

Thus, $\| Sf \| = \frac{1}{\sqrt{2}} \| f \|$, hence $\| S \| = \frac{1}{\sqrt{2}}$.

(ii) To show that $T^{*} = 2S$ it is necessary and sufficient to make sure that $(Tf, g) = (f, 2Sg)$ for all $f, g \in L^{2}(0,\infty)$, where $(\cdot, \cdot)$ denotes the scalar product in the Hilbert space $L^{2}(0,\infty)$. Indeed,

Here we use the substitution $t = 2s$.

To determine $S^{*}$ note that $S = \frac{1}{2} T^{*}$, therefore $S^{*} = \frac{1}{2} T^{**} = \frac{1}{2} T$.

Answer

(i) $\frac{1}{\sqrt{2}}$.

(ii) $S^{*} = \frac{1}{2} T$.