norm of a function in Hilbert space

Question

Let $L^2([0,1]) $ be the Hilbert space of all real valued square integrable functions on $[0,1]$ with the usual inner product. let $\phi$ be the linear functional $$\phi(f)=\int_{1/4}^{3/4}3\sqrt 2f d\mu$$, where $\mu$ denotes the Lebesgue measure on $[0,1]$. Then $||\phi||$ is?? My answer: $||\phi(f)||\leq \dfrac{3}{\sqrt 2}||f||\;\;\implies ||\phi||\leq \dfrac{3}{\sqrt 2}$ by using the definition of norm. Now if this inequality becomes an equality for a particular $f$, then is will be the norm, taking $f=1$, i.e a constant function, we get $$||\phi||=\dfrac{3}{\sqrt 2}$$ Is this right??

Answer 1

Is your functional $\varphi(f) = \int_{1/4}^{3/4} 3 \sqrt{2} f \, \mathrm{d} \mu$ or $\varphi(f) = \int_{1/4}^{3/4} \frac{3}{\sqrt{2}} f \, \mathrm{d} \mu$?

However, in both cases your answer is wrong. You have to use the corresponding norms! Hint: Since the dual space of $L^2([0,1])$ is isometric isomorphic to $L^2([0,1])$, we know that the functional $\phi(f) := \int g f \, \mathrm{d} \mu$, where $g \in L^2([0,1])$ has operator-norm $\|\phi\| = \|g\|_{L^2}$.