Exercise 2, § 2

Let $X =]0,1[$ and $\mu$ the Lebesgue measure on X. We define a sequence $(f_n)_{n\geqslant 1}$ of functions $X\rightarrow \mathbf{R}$ by setting

2 sin(2$n\pi x$) if $0< x <1/2$

$f_{2n}(x) =-2^n$ if $1-2^{-2n}< x <1-2^{-2n-1}$

0 otherwise,

and

2 sin(2$n\pi x$) if $0< x <1/2$

$f_{2n-1}(x) =2^n$ if $1-2^{-2n}< x <1-2^{-2n-1}$

0 otherwise,

for all $n\geqslant 1$ and all $x\in X$.

a) The family $(f_n)_{n\geqslant 1}$ is orthonormal in $\mathscr{L}^2(X, \mu)$.

b) For all $(x, y)\in X\times X$, set

Now prove that $|N(x, y)|\leqslant 2^{3/2}$ for all $(x, y)\in X\times X$. Deduce that $N_y\in \mathscr{L}^{\infty}(X)$ for all $y\in X$ and that

c) Now prove that, by setting

for $f\in \mathscr{L}^1(X, \mu)$ and $y\in X$, one defines by passing to quotients a continuous linear mapping from $L^1(X, \mu)$ into $L^2(X, \mu)$.

d) The linear mapping $u_N$ is not compact. (Consider the sequence $g_n=$ $2^n(f_{2n-1}-f_{2n})$; show that $g_n$ is integrable and that the sequence $(u_N(g_n))$ in $L^2(X, \mu)$ has no convergent subsequence.)