Norm of a linear functional in $L^2[0, 10]$.

Question

Let $L^2[0, 10]=\{f:[0, 10]\to\mathbb{R}\ |\ f$ is Lebesgue measurable and $\int_0^{10}f^2\, dx$ is finite$\}$ equipped with norm $\|f\|^2=\int f^2\, dx$ and let $T$ be a linear functional on $L^2[0, 10]$ given by $$T(f)=\int_0^2 f(x)\, dx-\int_3^{10}f(x)\, dx.$$ Then show that $\|T\|=3.$

Please tell how to start?

Answer 1

let us first write $$T(f)=\int_0^{10}g(x)f(x)\, dx=\langle g(x), f(x)\rangle,$$ where $g(x)$ is to be calculated. From above inner product, it is easy to see that $$\|T\|=\|g\|.$$So, the problem is to determine $g$. Observe that for $g(x)=\begin{cases} 1\ \mbox{for}\ 0\leq x\leq 2\\ 0\ \mbox{for}\ 2< x< 3\\ -1\ \mbox{for}\ 3\leq x\leq 10\end{cases}$ the required condition is satisfied. Further, $$\|g\|^2=\int_0^2 1\, dx+0+\int_3^{10}(-1)^2\, dx =9$$ which means that $\|T\|=\|g\|=3.$