Let $L^2([0, 1])$ be the Hilbert space of all real valued square integrable function on $[0, 1]$ with the usual inner product. Let $\phi$ be the linear functional on $L^2([0, 1])$ defined by
$$\phi(f) = \int_{\frac{1}{4}}^{\frac{3}{4}}3\sqrt 2 f d\mu $$
where $\mu$ denotes the Lebesgue measure on $[0, 1]$. Then $\vert \vert \phi \vert \vert=?$
How do I proceed? Please help.
Hint : $\|\phi\|=\sup_{\|f\|_2\leq1}|\phi(f)|$ and $$\int_{\frac{1}{4}}^{\frac{3}{4}}|f|d\mu \leq \Big(\int_{\frac{1}{4}}^{\frac{3}{4}} d\mu\Big)^{\frac12}\Big(\int_{\frac{1}{4}}^{\frac{3}{4}}f^2d\mu\Big)^{\frac12}=\frac{1}{\sqrt{2}} \Big(\int_{\frac{1}{4}}^{\frac{3}{4}}f^2d\mu\Big)^{\frac12}\leq \frac{1}{\sqrt{2}} \Big(\int_{\mathbb{R}}f^2d\mu\Big)^{\frac12}\leq \frac{1}{\sqrt{2}} $$ if $\|f\|_2\leq1$. In this case, you will have $$ |\phi(f)|\leq 3\sqrt{2}\int_{\frac{1}{4}}^{\frac{3}{4}}|f|d\mu\leq 3 $$ You need to show such a function $f$ verifying $|\phi(f)|=3$ or, at least a sequence $f_n$ such that $|\phi(f_n)|\rightarrow 3$