I need a some guidance through this particular problem: let $V$ be the space of functions $f:R \to R$ such that $\int_Rf^2(x)dx<+\infty$ and $\int_Rx^2f^2(x)dx<+\infty$
Consider on $V$ the product $\langle f,g \rangle:=\int_Rf(x)g(x)dx+\int_Rx^2f(x)g(x)dx$ which induces on $V$ the norm $||f||_V:=\langle f,f\rangle^{1/2}$
Finally, let $T: V \to R$ the linear operator defined as: $$T(f):=\int_Rfdx$$
The questions are to prove that $T$ is bounded and to calculate its norm. I have a general idea of the direction towards which I should move, but I'm unsure on how and when to use the data I've been given. I'm trying to understand the reasoning behind what I should be doing, and this is what I have:
To prove that $T$ is bounded (I think) I need to show that $|T(f)|_R<\infty$
We have $$|T(f)|_R=\bigg| \int_Rf(x)dx \bigg|$$
Now, I'm pretty sure I can write $$\bigg| \int_Rf(x)dx \bigg| \le\int_R|f(x)|dx$$ even if, in all honesty, I don't remember where this is from. Now, I know from the data I've been given that the norm inside that integral is $$\langle f, f\rangle^{1/2}=\biggl(\int_Rf^2(x)dx+\int_Rx^2f^2(x)dx \bigg)^{1/2}=\bigg( \int_R(1+x^2)f^2(x)dx \bigg)^{1/2}$$ on $V$.
And that's where I get stuck. I don't know where to go from here.
Thanks in advance.
$|\int f(x) dx| \leq \int \frac 1 {\sqrt {1+x^{2}}} (|f(x)|\sqrt {1+x^{2})}dx$. By Holder's inequality this gives $|Tf| \leq (\int \frac 1 {1+x^{2}} dx)^{1/2} \|f\|$ so $\|T\| \leq \sqrt {\pi}$. To show tht this value is attained on the unit disk just take $f(x)=\frac 1 {\sqrt {\pi} \sqrt {1+x^{2}}}$. Hence $\|T\| = \sqrt {\pi}$.