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Closed Graph Theorem

dray

Member
Feb 5, 2012
37
Suppose that 1<p<inf and a=(a_k) a complex sequence such that, for all x in l_p, the series (which runs from k=1 to inf) Sigma(a_k x_k) is convergent. Define T:l_p--->s by

Tx=y, where y_j=Sigma(a_k x_k) (where j runs from 1 to j).

I need to prove that

1) T has a closed graph (as a linear mapping from l_p to l_inf)

2) If a is the sequence defining T, then necessarily a in l_q, where (1/p)+(1/q)=1

Could somebody help me out with this please?
 

girdav

Member
Feb 1, 2012
96
What is $s$ here?
 

dray

Member
Feb 5, 2012
37
Here, s denotes the space of all possible sequences.
 

girdav

Member
Feb 1, 2012
96
Here, s denotes the space of all possible sequences.
Thanks.

In fact I don't understand the order of the question. Once we have proved that $a\in\ell^q$ (for example by the principle of uniform boundedness, or by contradiction, exhibiting $x\in\ell^p$ such that $\sum_{j=1}^{+\infty}|x_ja_j|=+\infty$), we can see that $T$ is continuous.
 

dray

Member
Feb 5, 2012
37
The question requires us to prove that T has a closed graph (i.e is continuous) first before we go on to show that a in l_q.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Suppose that $1<p<\infty$ and $a=(a_k)$ a complex sequence such that, for all $x$ in $l_p$, the series (which runs from $k=1$ to $\infty$) $\sum (a_k x_k)$ is convergent. Define $T:l_p\to s$ by

$Tx=y$, where $y_j=\sum (a_k x_k)$ (where $k$ runs from 1 to $j$).

I need to prove that

1) $T$ has a closed graph (as a linear mapping from $l_p$ to $l_\infty$)

2) If $a$ is the sequence defining $T$, then necessarily $a \in l_q$, where $(1/p)+(1/q)=1$.

Could somebody help me out with this please?
For 1), you have to show that if $x^{(n)}\to x$ in $l_p$ and $Tx^{(n)}\to y$ in $l_\infty$, then $Tx=y.$ To see that this condition holds, all you need to know is that, in each of $l_p$ and $l_\infty$, convergence in norm implies coordinatewise convergence. The $j$th coordinate of $Tx^{(n)}$ is $$(Tx^{(n)})_j = \sum_{k=1}^j a_kx^{(n)}_k$$ (where $x^{(n)}_k$ denotes the $k$th coordinate of $x^{(n)}$).

Let $n\to\infty$ on both sides of that displayed equation to see that $y_j = (Tx)_j$. Since that holds for each $j$ it follows that $Tx=y$, as required.

For 2), fix $n$ and define $z^{(n)}\in l_p$ by $$z^{(n)}_k = \begin{cases}\theta_k a_k^{q-1}&\text{if }1\leqslant k\leqslant n, \\0&\text{if } k>n, \end{cases}$$ where $\theta_k$ is a complex number of absolute value 1 chosen so as to make $\theta_k a_k^q$ real and nonnegative. Then $$\|z^{(n)}\|_p = \Bigl(\sum_{k=1}^n |a_k|^{p(q-1)}\Bigr)^{1/p}.$$ The $n$th coordinate of $Tz^{(n)}$ satisfies $$Tz^{(n)}_n = \sum_{k=1}^n a_kz^{(n)}_k = \sum_{k=1}^n |a_k|^q.$$

But it follows from 1) and the closed graph theorem that $T$ is bounded, and therefore $\|Tz^{(n)}\|_\infty \leqslant \|T\|\,\|z^{(n)}\|_p.$ You should be able to deduce that $$\Bigl(\sum_{k=1}^n |a_k|^q\Bigr)^{1/q} \leqslant \|T\|.$$ Finally, let $n\to\infty$ to conclude that $a\in l_q$.
 

dray

Member
Feb 5, 2012
37
How do I prove that for each sequence space, convergence in norm implies coordinatewise convergence?
 

girdav

Member
Feb 1, 2012
96
Because for each norm and each sequence $x=\{x_k\}$, we have $|x_k|\leq \lVert x\rVert$.
 

dray

Member
Feb 5, 2012
37
Because for each norm and each sequence $x=\{x_k\}$, we have $|x_k|\leq \lVert x\rVert$.
Thanks girdav, but I'm still looking for a proof that this is the case.
 

girdav

Member
Feb 1, 2012
96
I don't understand, you want a proof of the last inequality? If it's the $\infty$ norm it's obvious, and for $p$-norms, just use the fact that the series which defines the norm is greater than only one term of this series.