Norm of a functional defined by an integral

Question

Show that $\phi: L^{\frac{3}{2}}\left(\left(0,\frac{1}{2} \right] \right) \to \mathbb{R}$ defined by

$$u\mapsto\int_0^{\frac{1}{2}}u \, dx$$

is linear. Find $\|\phi\|_{\mathcal{L}}$.

My try

Clearly $\forall u, v\in L^{\frac{3}{2}}\left(\left(0,\frac{1}{2}\right]\right)$ and $\lambda\in\mathbb{R}$

1. $\\ \phi(u+v)=\int_{0}^{\frac{1}{2}}(u+v)dx=\\ =\int_{0}^{\frac{1}{2}} u \, dx+\int_0^{\frac{1}{2}} v \, dx=\\ =\phi(u)+\phi(v)$

2. $\phi(\lambda u)=\int_{0}^{\frac{1}{2}}\lambda u \, dx = \lambda \int_0^{\frac{1}{2}} u \, dx=\lambda\phi(u)$

so the linearity of the functional is easy to prove (right?), but what can I do to evaluate its norm? Maybe it can be useful the Holder Inequality? I think it's not the right idea. Any help would be appreciated.

Answer 1

We have $$ |\phi u|\le\int_0^{1/2}|u|\,dx\le\left(\int_0^{1/2}|u|^{3/2}\,dx\right)^{2/3}\left(\int_0^{1/2}dx\right)^{1/3} = 2^{-1/3}\|u\|_{L^{3/2}}. $$ So $\|\phi\|\le 2^{-1/3}$. To show equality, let $u(x) = 2^{2/3}$, $x\in [0,\tfrac 1 2]$. Then $$ \|u\|_{L^{3/2}} = \left(\int_0^{1/2}|u|^{3/2}\,dx\right)^{2/3} = \left(\int_0^{1/2}2\,dx\right)^{2/3} = 1 $$ and $$ \phi u = \int_0^{1/2}u(x)\,dx = \frac 1 2\cdot 2^{2/3} = 2^{-1/3}. $$