Let $L_1([0,1], m)$ be the Banach space of $\mathbb{K}$-valued (i.e. $\mathbb{C}$ or $\mathbb{R}$-valued) Lebesgue-integrable functions, where $m$ is the Lebesgue measure, be equipped with the norm $‖f‖_1=\int_{[0,1]}|f|\,dm$ for $f \in L_1([0,1], m)$. For $n \geq 1$ define $g_n(x) =n\sin(n^2x)$, for $x \in [0,1]$.
Furthermore, let $\phi_n(f)=\int_{[0,1]} f g_n \, dm$.
I'm then asked to show that $\phi_n(f)$ defines a bounded linear map (which I've done) and to show $‖\phi_n‖=n$ for $n \geq 2$. What I've done is that I have shown $‖\phi_n‖\leq n$. If I could find a function $f \in L_1([0,1], m)$ such that $|\phi_n(f)|=n$ and $‖f‖_1=1$, then I would have shown equality. How can I find such a function?
The problem also asks whether I can find a function $f \in L_1([0,1], m)$ such that $\lim_{n \rightarrow \infty} |\phi_n(f)|=\infty$ (perhaps I could use the same function as above?)
Best Answer
It is a general fact that if $g$ is a bounded function, then the norm of the linear functional $L\colon \mathbb L^1\to\mathbb R$ defined by $L(f)=\int_{(0,1)}g(x)f(x)\mathrm d\lambda(x)$ is $\left\lVert g\right\rVert_\infty$. Indeed, fix a positive $\varepsilon<\left\lVert g\right\rVert_\infty$ and define the sets $A^+:=\left\{x\in (0,1), g(x)\gt \left\lVert g\right\rVert_\infty-\varepsilon\right\}$ and $A^-:=\left\{x\in (0,1), -g(x)\gt \left\lVert g\right\rVert_\infty-\varepsilon\right\}$. Then $$ \max\left\{L\left(\frac 1{\lambda\left(A^+\right)}\mathbf 1_{A^+}\right),L\left(\frac 1{\lambda\left(A^-\right)}\mathbf 1_{A^-}\right) \right\}\geqslant \left\lVert g\right\rVert_\infty-\varepsilon $$ and (at least one of) the involved function are well-defined and their $L^1$-norm is $1$.
For the other question, look at one of the functions $x\mapsto \pm x^{-\alpha}$ for $1/2\lt\alpha\lt 1$.