This is another (at least to me!) difficult question I found that I would appreciate some help/insight on:
If B = a real Banach space which is just $C[0,1]\mapsto R$ with elements y = y(x)
and $H$ is a real Hilbert Space $L_{2}[-1,1]$ which is just the completion of the space of continuous functions $C[-1,1]$ with L2 norm $(\|{v}\|)^{2} = \int_{0}^{1}v^{2}(x)dt$
Question: given $g(x)$, a continuous function on domain $[0,1]$ and $G$ is just multiplication by $g(x)$, how would I find the norm of $G$ in these two spaces?
I have a general idea that I need to use the operator norm on the first and the l2 norm on the second, but I'm not sure how to actually compute these.
Thank you
So the norm of the operator $G$ can be defined by $$ \|G\|_{B\to B} = \sup_{\varphi\in B} \frac{\|G\varphi\|_{B}}{\|\varphi\|_{B}}. $$ If $B=C^0([0,1])$, the norm is the $L^\infty$ norm. We can compute for any $\varphi ∈ C^0([0,1])$ the norm of $$ \|G\varphi\|_{B} = \|g\varphi\|_{L^\infty} ≤ \|g\|_{L^\infty}\|\varphi\|_{L^\infty} = \|g\|_{L^\infty}\|\varphi\|_{B} $$ Therefore, $\|G\| ≤ \|g\|_{L^\infty}$. But we can take $g=\varphi$ and see that we have equality in that case. Therefore, $\|G\| = \|g\|_{L^\infty}$. The same work if $B = L^2$.