Exercise 9, § 1

For every pair $(x, y)\in ]0,1]^2$, there exists an integer $n\geqslant 1$ and an integer $k$ such that $0\leqslant k\leqslant 2^{n-1}-1$ such that

2 2

We then define $N : ]0,1]\times ]0,1]\rightarrow \mathbf{C}$ by setting

$\{2$ if $\frac{2k}{2^n}< y\leqslant \frac{2k+1}{2^n}$

0 if $\frac{2k+1}{2^n}< y\leqslant \frac{2k+1}{2^n}$.

a) Now prove that the formula

defines a continuous linear mapping $u$ on $L^1([0,1])$ with respect to Lebesgue measure.

b) Now prove that the endomorphism $u$ is not compact.

c) The endomorphism $u^2$ is compact.

d) Let $(x_n)_{n\in\mathbf{N}}$ be a bounded sequence in $L^1([0,1])$. The sequence $(u(x_n))_{n\in\mathbf{N}}$ admits a weakly convergent subsequence.

e) Let $(x_n)_{n\in\mathbf{N}}$ be a weakly convergent sequence in $L^1([0,1])$. The sequence $(u(x_n))_{n\in\mathbf{N}}$ converges in $L^1([0,1])$.

f) Let E be the topological vector space whose elements are the functions $f: [0,1]\rightarrow \mathbf{C}$ which are bounded and measurable, endowed with the finest topology which is stronger than the topology of $L^2([0,1])$ and than the weak topology relative to $L^1([0,1])$. Then the linear mapping $v$ defined by

is a compact endomorphism of E. Its transpose $^tv$ is not compact for the strong topology on $E'$.