A bounded linear operator

[gugo82]

Let $l^1={x=(x_n) subseteq RR:quad \sum_(n=1)^(+oo)|x_n|<+oo}$, $c_0={y=(y_n)subset RR:quad lim_n y_n=0}$ and $l^oo={x^**=(x_n^**) subset RR:quad "sup"_(n in NN) |x_n^**| <+oo}$.

The two sets of real sequences $l^1, c_0$ turn out to be Banach spaces with their canonical real vector space structure and the norms defined by:

$AA x in l^1,quad ||x||_1=\sum_(n=1)^(+oo)|x_n|$,

$AA y in c_0,quad ||y||_oo="sup"_(n in NN) |y_n|$.

The normed duals $(l^1)^**$ and $(c_0)^**$ are, respectively, $(l^oo,||\cdot ||_oo)$ (the $oo$-norm is the one defined above) and $(l^1,||\cdot ||_1)$: their both Banach spaces.
(No need to proof; we recall these notions and take them as a prerequisite).

Exercise:

Prove that the function $T$ defined by:

$AA x=(x_n) in l^1, quad Tx=(1/n*sum_(k=1)^nx_k)$

1) is a bounded linear operator of $l^1$ in $c_0$;

2) is one-to-one, that is $N(T)={x in l^1:quad Tx=0}={0}$ (here $0$ is the null sequence);

3) find a $y in c_0$ s.t. $y \notin R(T)=T(l^1)$ and prove that $c_(00)={y in c_0:quad (exists mu in NN: AA nge mu, y_n=0)} subseteq R(T)$;

4 a) prove that $R(T)$ is not closed in $c_0$ (to be more precise, $R(T)$ is everywhere dense in $c_0$) using your answer to question 3);

4 b) use the following characterization of one-to-one closed range operators to answer question 4 a) by reductio ad absurdum without using 3):

Let $(X,||\cdot||_X),(Y,||\cdot||_Y)$ be Banach spaces and $T in L(X,Y)$.
$T$ is a one-to-one closed range operator iff $\exists C>0: quad AA x in X, ||x||_Xle C*||Tx||_Y$.

5) Find the adjoint operator of $T$, that is the unique operator $T^**:l^1=(c_0)^** to (l^1)^**=l^oo$ s. t.:

$AA y^** in l^1, AA x in l^1, quad \sum_(n=1)^(+oo) (T^**y^**)_nx_n= = = \sum_(n=1)^(+oo) y^**1/n\sum_(k=1)^nx_k$.

[elgiovo]

I try with number 1:

a) $T$ is bounded: $Tx=1/n sum_(k=1)^n x_k<1/n sum_(k=0)^(oo)|x_k|=1/n ||x||_1<+oo$;

b) $T$ is linear: $T(alphax+betay)=sum_(k=1)^n(alpha x_k+beta y_k)=alpha sum_(k=1)^nx_k+beta sum_(k=1)^ny_k=alphaTx+betaTy$;

c) $T:l^1 to c_0$: we need to evaluate $lim_(n to oo) 1/n sum_(k=1)^n x_k$. Define the function $X(t)={(0" ",t<0),(x_k cdot chi_{[k,k+1]}" ",t>=0):}$.
Then $lim_(n to oo) 1/n sum_(k=1)^n x_k=lim_(n to oo) 1/n int_(-n)^n X(t)dt$, which is the average of a $ccL^1$ function, i.e. $0$.