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.:

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

[Gabriel6]

"Gugo82":
As elgiovo noted before, it's $||Tx||_oole||x||_1$ so that $||T||_(op)le 1$. Say $e^1=(delta_n^1)$, $delta_n^1$ being the Kronecker delta: we have $Te^1=(1/n)$ so that $||Te^1||_oo=1=||e^1||_1$; [...] so $||T||_(op)=1$.

Ok.

"Gugo82":If $T$ has an eigenvalue $lambda in CC$, then $lambda=1/nu$ for one $nu in NN$ [...] But none of the number $1/nu$ can actually be an eigenvalue of $T$. Let's suppose $1/nu$ is an eigenvalue and let's try to calculate an eigenvector $x$: as we noted before, all the coordinates of $x$ with indexes $n
Actually, $z = 0$ isn't possible, but this isn't really significant -- ok.

[gugo82]

"Gabriel":6. Now that we know $T$ is bounded in the operator norm $||\cdot||_{op}: l^1 \to c_0$, find $||T||_{op}$.

As elgiovo noted before, it's $||Tx||_oole||x||_1$ so that $||T||_(op)le 1$.
Say $e^1=(delta_n^1)$, $delta_n^1$ being the Kronecker delta: we have $Te^1=(1/n)$ so that $||Te^1||_oo=1=||e^1||_1$; this relation shows that $||T||_(op)$ can't actually be $<1$ so $||T||_(op)=1$.

"Gabriel":7. Find any $\lambda \in CC$ for which there exists a non-zero sequence $x \in l^1$ such that $Tx = \lambda x$.

If $T$ has an eigenvalue $lambda in CC$, then $lambda=1/nu$ for one $nu in NN$: infact there exists $x=(x_n) in l^1-{0}$ s.t. $\sum_(k=1)^n x_k=n*lambda*x_n$ or better s.t.:

(*) $quad \{((1-lambda)*x_1=0, " for " n=1),(\sum_(k=1)^(n-1)x_k + (1-n*lambda)*x_n=0, " for " nge 2):} quad$;

if $x_nu$ is the first non-zero coordinate of $x$, then the homogeneous linear system:

(**) $quad \{((1-lambda)x_1=0),(x_1+(1-2lambda)x_2=0),(\ldots),(x_1+\ldots+x_(nu-1)+(1-nulambda)x_nu=0):} quad $,

obtained by taking the first $nu$ equation of (*), has non-trivial solutions: this implies $lambda=1/nu$, because system (**) is lower triangular and $x_1,x_2,\ldots, x_(nu-1)=0$.

But none of the number $1/nu$ can actually be an eigenvalue of $T$. Let's suppose $1/nu$ is an eigenvalue and let's try to calculate an eigenvector $x$: as we noted before, all the coordinates of $x$ with indexes $nnu$, the coordinate $x_(nu+p)$ has to satisfy the recurrence equation $((nu+p)/nu-1)*x_(nu+p)=\sum_(k=1)^(nu+p-1)x_k=(nu+p-1)/nu*x_(nu+p-1)$, which leads to the closed form:

(***) $quad x_(nu+p)=((nu+p-1)!)/((nu-1)!p!)*z quad$;

relation (***) also shows that $|x_(nu+p)|ge |z|$ (because $((nu+p-1)!)/((nu-1)!p!)ge1$).
Now we got a contradiction: infact, if $z=0$ then $||x||_1=0$ else $||x||_1=\sum_(k=nu)^(+oo)|x_k|=|z|*{1+\sum_(p=1)^(+oo)((nu+p-1)!)/((nu-1)!p!)}ge |z|*\sum_(p=0)^(+oo)1=+oo$; then $||x||_1=oo$ or $||x||_1=0$, in contrast with $x in l^1-{0}$ (remember the definition of eigenvalue).

[Gabriel6]

6. Now that we know $T$ is bounded in the operator norm $||\cdot||_{op}: l^1 \to c_0$, find $||T||_{op}$.

7. Find any $\lambda \in CC$ for which there exists a non-zero sequence $x \in l^1$ such that $Tx = \lambda x$.

8. Is $T$ compact?

[gugo82]

"Gabriel":It seems I forgot some $k$ while posting - I'm going to edit.

Correct!

[Gabriel6]

It seems I forgot some $k$ while posting - I'm going to edit.

[gugo82]

"Gabriel":5 is just the wind of a moment whispering mysterious words to the ears of your mind. Let $y \in l^1 = c_0^\star$ and $u = T^\star y$. We postulate that $u_n = \sum_{k=n}^\infty y_k$ - and it's a trivial exercise to check the guess is right. []

Wrong answer.
You forgot something...

Try with a direct calculation... and don't venture a guess this time!

[Gabriel6]

5 is just the wind of a moment whispering mysterious words to the ears of your mind. Let $y \in l^1 = c_0^\star$ and $u = T^\star y$. We postulate that $u_n = \sum_{k=n}^\infty \frac{y_k}{k}$ - and it's a trivial exercise to check the guess is right. []

[Gabriel6]

"Gugo82":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.

Please note that, even if without ever mentioning it, I proved the statement in the quote precisely using that $c_{00}$ is a subspace of $R(T)$ and everywehere dense in $c_0$.

"Gabriel":4.a is intriguing. Provided $y \in c_0$, we need to prove that, for any $\epsilon > 0$, there exists $x \in l^1$ such that $||y - Tx||_\infty < \epsilon$. The idea definitively lacks of fantasy - we construct $x$ by recurrence. As $y \in c_0$, start getting $v \in NN$ such that $|y_n| < \epsilon$, for any $n > v$. So consider the (triangular) linear system of equations $\frac{1}{k} \sum_{i=1}^k x_i = y_k$ ($k = 1, 2, \ldots, v$) and observe it is full-rank - that granting it has one solution $(x_1, x_2, \ldots, x_v) \in CC^v$. Then let $x_{v+1} = -\sum_{k=1}^v x_k$ and $x_n = 0$, for any $n \ge v+2$. Obviously, $x = \{x_n\}_{n \in NN} \in l^1$. Moreover, setting $z = Tx$ gives that $z_n = y_n$, for any $n = 1, 2, \ldots, v$, and $z_n = 0$, whenever $n > v$ (by construction). Hence $||y - Tx||_\infty =$ sup$\{|y_n - z_n|\}_{n \in NN} = $sup$\{|y_n|\}_{n > v} < \epsilon$. Then $R(T)$ is dense in $c_0$. But it's not closed in $c_0$, as 3 actually proves. []

[Gabriel6]

"Gugo82":3) find a $y in c_0$ s.t. $y \notin R(T)=T(l^1)$ and prove that $c_(00)={y in c_0:quad (exists mu in NN: AA nge mu, y_n=0)} subseteq R(T)$.

In fact, I didn't even notice the second part when I read the problem - no pain! Let $y \in c_{00}$. Then there exists $v \in NN$ such that $y_n = 0$, whenever $n \ge v$. So set $(x_1, x_2, \ldots, x_v)$ to the (unique) solution of the (triangular) full-rank linear system of equations $\frac{1}{k} \sum_{i=1}^k x_i = y_k$ ($k = 1, 2, \ldots, v$); $x_{v+1} = - \sum_{i=1}^v x_i$ and $x_n = 0$, for any $n > v+1$. By construction, $x = \{x_n\}_{n \in NN}\in l^1$ and $Tx = y$. So $y \in R(T)$ - i.e., $c_{00} \subseteq R(T)$. []

[gugo82]

"Gabriel":3 is funny. For any $n \in NN = \{1, 2, \ldots\}$, let $y_n = 0$, if $n$ is odd, and $y_n = \frac{1}{n}$, if $n$ is even. Clearly $\lim_{n \to \infty} y_n = 0$ - i.e., $y \in c_0$. Now suppose there exists $x \in l^1$ such that $Tx = y$. Then $x_1 = y_1 = 0$ and $-x_{2n+1} = x_{2n} = 1$, for any $n \in NN$ (by induction). Hence $\lim_{n \to \infty} |x_n| = 1$ - which is absurd, as $x \in l^1$.

Ok, your $y in c_0$ isn't in the range of $T$, but you didn't succeed to show $c_(00) subset R(T)$ (the proof of this assertion is trivial).

These two facts, together with the relation $bar(c_(00))=c_0$ in $||\cdot ||_oo$, can be used in a reductio ad absurdum in order to answer question 4 a).
In fact, let's suppose that $R(T)$ is closed in $c_0$: then $c_(00) subset R(T)$ implies $c_0 =bar(c_(00)) subseteq bar(R(T))=R(T) subseteq c_0$ and so $R(T)=c_0$, which is an absurd because $T$ is not surjective.
Then $T$ is not of closed range; to be more precise, $R(T)$ is everywhere dense in $c_0$ because $c_(00) subset R(T)$.


P.S.: I slightly modified the text of questions 4 a,b).

[gugo82]

"elgiovo":[quote="Gugo82"]

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 $\exists C>0: quad AA x in X, ||x||_Xle C*||Tx||_Y$.

A corollary of Banach-Steinhaus' theorem says the right implication is true:
$X$, $Y$ Banach, $T:X to Y$ linear, continuous, one-to-one $to$ $exists C>0$ $|$ $||x||_X<=C||Tx||_Y$.[/quote]
That's right! The right implication's an almost trivial consequence of B-S' Theorem.

The left implication is also true.

In fact, suppose that there exists $C>0$ s.t. $||x||_Xle C*||Tx||_Y$ holds for all $x in X$: then $x in N(T) quad hArr quad x=0_X$ and so $T$ is one-to-one (or injective).
In order to prove that $R(T)=bar(R(T))$ in $Y$, let's fix a $y in bar(R(T))$ and show that $y in R(T)$: we call $(y^m)$ a sequence of elements in $R(T)$ which converges to $y$ (such a sequence exists because the norm topology on $Y$ is Hausdorff's) and name $(x^m)$ the only sequence in $X$ s.t. $AA m in NN, y^m=Tx^m$ (such a sequence exists because $T$ is iniective and $(y^m) subset R(T)$). Now we have:

$AA m,p in NN, quad ||x^(m+p)-x^m||le C*||T(x^(m+p)-x^m)||_Y=C||y^(m+p)-y^m||_Y$

so $(x^m)$ is a Cauchy's in $X$ because $(y^m)$ is a Cauchy's in $Y$. The completeness of $X$ implies that exists $x in X$ s.t. $x^m to x$ in norm; then we get also $y^m=Tx^m to Tx$ by continuity. But a sequence which converges in $Y$ cannot have two distinct limits cause $Y$ is Hausdorff's: then $Tx=y$ and $y in R(T)$.
The double inclusion $bar(R(T)) subseteq R(T) subseteq bar(R(T))$ holds and so $T$ has closed range.

[elgiovo]

"Gabriel":So what's wrong with my previous example? Apart from a typo on a sign that I am going to fix.

Your example is Ok. I was only trying to correct Gugo's statement.

[Gabriel6]

"Gugo82":
There's a simpler answer. $Tx in c_0$ because of Cesaro's Theorem on arithmetic

There is even a simpler one - as $x \in l^1$, then $0 \le s=$sup$\{|x_n|\}_{n \in NN} < \infty$ and indeed $s = \max\{|x_n|\}_{n \in NN}$. Hence $|\frac{1}{n} \sum_{k=1}^n x_k| \le \frac{1}{n} \sum_{k=1}^n |x_k| \le s \le \sum_{n=1}^\infty |x_n| = ||x||_1$, for any $n \in NN$. So $||Tx||_\infty \le ||x||_1$. []

[Gabriel6]

"elgiovo":
A corollary of Banach-Steinhaus' theorem says the right implication is true:
$X$, $Y$ Banach, $T:X to Y$ linear, continuous, one-to-one $to$ $exists C>0$ $|$ $||x||_X<=C||Tx||_Y$.

So what's wrong with my previous example? Apart from a typo on a sign that I am going to fix.

[gugo82]

@elgiovo:

"elgiovo":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. $0$.

There's a simpler answer.
$Tx in c_0$ because of Cesaro's Theorem on arithmetic means:

If a sequence $(x_n)$ converges in $RR$ then the sequence of its arithmetic means, i.e. the one of $n$-th term $1/n\sum_(k=1)^nx_k$, converges to the same limit.

[elgiovo]

"Gugo82":

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 $\exists C>0: quad AA x in X, ||x||_Xle C*||Tx||_Y$.

A corollary of Banach-Steinhaus' theorem says the right implication is true:
$X$, $Y$ Banach, $T:X to Y$ linear, continuous, one-to-one $to$ $exists C>0$ $|$ $||x||_X<=C||Tx||_Y$.

[Gabriel6]

"Gugo82":4 b) use the following characterization of one-to-one closed range operators to answer question 4 a) directly:

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 $\exists C>0: quad AA x in X, ||x||_Xle C*||Tx||_Y$.

4.b is impossible - as the property you quoted is false. In fact, fix $n \in NN = \{1, 2, \ldots\}$ and let $-x_{2k-1} = x_{2k} = \frac{1}{k}$, for any $k = 1, 2, \ldots, n$, and $x_k = 0$, whenever $k > 2n$. Clearly, $x = \{x_k\}_{k \in NN} \in l^1$ and $||x||_1 = 2\cdot \sum_{k=1}^n \frac{1}{k}$. On the other hand, setting $y = Tx$ yields that $y_k = \frac{x_k}{k}$, if $k$ is odd, and $y_k = 0$, if $k$ is even - which proves $||y||_{\infty} = |y_1| = 1$. But we can choose $n \in NN$ in such a way that $||x||_1 > C$, for any given $C > 0$. []

[Gabriel6]

4.a is intriguing. Provided $y \in c_0$, we need to prove that, for any $\epsilon > 0$, there exists $x \in l^1$ such that $||y - Tx||_\infty < \epsilon$. The idea definitively lacks of fantasy - we construct $x$ by recurrence. As $y \in c_0$, start getting $v \in NN$ such that $|y_n| < \epsilon$, for any $n > v$. So consider the (triangular) linear system of equations $\frac{1}{k} \sum_{i=1}^k x_i = y_k$ ($k = 1, 2, \ldots, v$) and observe it is full-rank - that granting it has one solution $(x_1, x_2, \ldots, x_v) \in CC^v$. Then let $x_{v+1} = -\sum_{k=1}^v x_k$ and $x_n = 0$, for any $n \ge v+2$. Obviously, $x = \{x_n\}_{n \in NN} \in l^1$. Moreover, setting $z = Tx$ gives that $z_n = y_n$, for any $n = 1, 2, \ldots, v$, and $z_n = 0$, whenever $n > v$ (by construction). Hence $||y - Tx||_\infty =$ sup$\{|y_n - z_n|\}_{n \in NN} = $sup$\{|y_n|\}_{n > v} < \epsilon$. Then $R(T)$ is dense in $c_0$. But it's not closed in $c_0$, as 3 actually proves. []

[Gabriel6]

3 is funny. For any $n \in NN = \{1, 2, \ldots\}$, let $y_n = 0$, if $n$ is odd, and $y_n = \frac{1}{n}$, if $n$ is even. Clearly $\lim_{n \to \infty} y_n = 0$ - i.e., $y \in c_0$. Now suppose there exists $x \in l^1$ such that $Tx = y$. Then $x_1 = y_1 = 0$ and $-x_{2n+1} = x_{2n} = 1$, for any $n \in NN$ (by induction). Hence $\lim_{n \to \infty} |x_n| = 1$ - which is absurd, as $x \in l^1$. []

[Gabriel6]

2 is completely trivial. In fact, let $y = Tx$ and suppose $Tx = 0$, provided $x \in l^1$. Maybe you'll think it's amazing, but $x_1 = 0$ - since $y_1 = 0$ and $y_1 = x_1$. Now fix $n \in NN$ and admit $x_k = 0$, for each $k = 1, 2, \ldots, n$. Then $y_{n+1} = 0$ gives $\sum_{k=1}^{n+1} x_k = 0$ - i.e., $x_{n+1} = 0$. So the claim follows by strong induction. []