Answer to Question #164916 in Functional Analysis for Ebenezer

Question

Let c0 be the space of sequences of complex numbers which converge to 0. That is

c_0 = {(x_i)_i∈N : xi ∈ C, x_i → 0}.

(i) Show that c0 is a closed subspace of L^∞.

(ii) Define a mapping T by

T : L^∞ → c_0

(x_n)_n → (x_n/ n ) _n

Show that T is a (linear bounded) operator. Show that ran T is not closed.

Answer 1

i) Let $x^{(k)}\isin c_0$ be a sequence converging to $w\isin L^{\infin}$. Take $\epsilon>0$ and $N_0\isin N$ such that

$\displaystyle \sup_{1\leq i\leq \infin}|x^{(k)}_i-w_i|<\frac{\epsilon}{2}$

for all $k>N_0$. For each $k$ choose $N_1\isin N$ such that

$|x^{(k)}_i|\leq \frac{\epsilon}{2}$ for all $i>N_1$ .

Thus, $|w_i|=|w_i-0|\leq |w_i-x^{(k)}_i|+|x^{(k)}_i|\leq \epsilon$ for $i>N_1$, and $k>N_0$ , that means that the sequence $w_i$ converges to and $w_i\isin c_0$ . Hence $c_0$ is closed in $L^{\infin}$ .

ii) For linear bounded) operator $T:V_1\to V_2$

$||Tv||_2\leq C||v||_1,$ constant $C>0$

In our case:

for $L^{\infin}: ||x||=sup|x_n|$

for $c_0$ : $||x||=sup|x_n/n|$

Then:

$sup|x_n/n|\leq Csup|x_n|$

So, there exists $C>0$ . That is, $T$ is a linear bounded operator.

$Ran(T)=(x_n/n)_n$

If sequence $(x_n)$ in $L^{\infin}$ does not converge, then sequence $(x_n/n)_n$ does not converge to 0.

So, $Ran(T)$ is not closed.