I've been taking functional analysis and trying to wrap my head around a question I've been asked for it.
Consider $X = L ^ 2 (−1, 1)$ with the inner product $\langle\cdot,\cdot\rangle\to \mathbb{R}$ given by $\langle x,y\rangle = \int_{-1}^{1}x(s)y(s)ds$
It induced norm $\Vert x\Vert = \sqrt{\langle x,x\rangle}$ fix $u$ in $L^2(1,-1)$ and let $v(t) = e^{-t}$ for any $t \in [-1,1]$. Let the operator $T: X\to X$ be defined as $Tx(t) = v(t) \langle x,u\rangle$.
Find $\Vert T\Vert$
I've so far got that $\sqrt{\langle Tx,Tx\rangle} = \sqrt{\int_{-1}^{1}(e^{-2s}\langle x,u\rangle)ds}$ with no further ideas on how to progress. I've tried using Riesz's theorem and the Cauchy-Schwartz inequality with no success. I vaguely understand that the goal is to show that $c\leq\Vert T\Vert\leq c$ for some $c$ which is the answer, but I don't have any clue how to get there. Any advice would be appreciated, especially if you can make me understand what's going on.
Thank you.
A quick remainder, the norm $\lvert \lvert T \rvert \rvert$ is defined in the following way $$\lvert \lvert T \rvert \rvert=\underset{x\neq 0\in L^{2}}{\sup}\frac{\lvert \lvert Tx \rvert \rvert}{\lvert \lvert x \rvert \rvert}=\underset{x\in L^{2},\lvert \lvert x \rvert \rvert=1}{\sup}\lvert \lvert Tx \rvert \rvert$$ First, you can prove one inequality, by using Cauchy-Schwartz inequality twice, it gives $$ \lvert \langle x,u \rangle \rvert \leq \lvert \lvert x \rvert \rvert_{2}\times \lvert \lvert u \rvert \rvert_{2}$$ Then, we can re-use Cauchy-Schwartz $$\lvert \langle e^{-s}\langle x,u\rangle,e^{-s}\langle x,u\rangle \rangle\rvert\leq \lvert \lvert x \rvert \rvert_{2}\times \lvert \lvert u \rvert \rvert_{2} \lvert \lvert v \rvert \rvert_{2} $$ So; it proves the upper bound on $\lvert \lvert T\rvert\rvert$.To prove the other bound, you can just take $u_0=\frac{u}{\lvert \lvert u \rvert \rvert}$.