Show that $A\in \mathcal{B}(X_{2},X_{1}).$ and Compute the norm $\Vert A\Vert.$

Question

Please leave some hint and comments to the following Problem not the whole solution

Problem: Let $X_{1}$ be the normed linear space $C([0,1],\mathbb{R})$ equipped with the norm $\Vert f\Vert_{1}=(\int_{0}^{1}\vert f(t)\vert)$, $X_{2}$ be the normed linear space $C([0,1],\mathbb{R})$ equipped with the norm $\Vert f\Vert_{2}=(\int_{0}^{2}\vert f(t)\vert)$. Given $f\in X_{2},$ let $A(f(t))=tf(t).$ Show that $A\in \mathcal{B}(X_{2},X_{1}).$ and Compute the norm $\Vert A\Vert.$

What I have done: $$\Vert A(f(t))\Vert_{1}=\int_{0}^{1}\vert tf(t)\vert dt\leq (\int_{0}^{1}t^{2}dt)^{\frac{1}{2}}(\int_{0}^{1}\vert f(t)\vert^{2}dt)^{\frac{1}{2}} $$ $$=\sqrt{\frac{1}{3}}\Vert f\Vert_{2}<\infty $$ So $A(f(t))\in X_{1}.$ And $\Vert A\Vert \leq \sqrt{\frac{1}{3}}. $

This completely seems logical to me but everything is fishy. Because if I calculate $f(t)=\sqrt{3}t$, then I get $\Vert A(f(t))\Vert=\frac{\sqrt{3}}{3}. $

Answer 1

Your answer is correct.

With $f(t)=\sqrt{3} t$ you have $\| f\|_2=\sqrt{3} \frac{1}{\sqrt{3}}=1$ and as you computed $\|A f\|_1=\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}}$ so: $$\| A f\| \leq \frac{1}{\sqrt{3}} \|f\|_2$$ and there is no problem.