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Here is the question:

Let $A = [a_{ij}]_{i,j = 1}^{\infty}$ be an infinite matrix of real numbers and suppose that, for any $x \in \ell^2,$ the sequence $Ax$ belongs to $\ell^2.$ Prove that the operator $T,$ defined by $T(x) = Ax,$ is a bounded operator on $\ell^2.$

My question is:

I got a hint to use the uniform boundedness principle here but I do not know why, could anyone explain this to me, please? what makes me when I look at a problem to decide that it should be solved by UBP?

EDIT:

1-I have taken this proposition: "The series $\sum_{n =1}^{\infty} a_{n} b_{n}$ converges absolutely for every convergent sequence $\{b_{n}\}$ iff $\sum_{n =1}^{\infty} |a_{n}|$ converges." will it be helpful here in our case? the problem is that here in our case we are in $l^2.$

2-Also, Should it be better to use the uniform boundedness principle or the following theorem to solve the problem given above?

Theorem:

Let $X,Y$ be Banach spaces and let $\{T_{n}\}_{n=1}^{\infty}$ and $T$ be operators in $\mathcal{L}(X,Y).$ then $\lim_{n} T_{n}x = Tx,$ for all $x \in X,$ iff

(a)the sequence $\{T_{n}\}$ is bounded;

(b)lim_{n} T_{n}x exists on a dense subset of $X.$

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    $\begingroup$ Let $A_n$ be the top-left $n\times n$ matrix (zeroing out everything else), then for each $x$, $\sup_n \|A_n x\|_2 < \infty$ and by the uniform boundedness principle, $\sup_n \|A_n\| < \infty$. Finally note that $Ax$ is the pointwise limit of $A_nx$. $\endgroup$
    – user58955
    Apr 6, 2020 at 3:09
  • $\begingroup$ @user58955 could you provide more details, please? Also, what about my question above? How do I know that I should apply UBP? $\endgroup$
    – user593471
    Apr 6, 2020 at 3:15
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    $\begingroup$ Such questions are usually hard to answer. There is really no protocol. So, you want to show that $\|Ax\|$ is uniformly upper bounded for all unit vectors $x$. Now you are dealing with infinite sums and you need to use the fact that $Ax\in \ell^2$ for all $x$. Thus you may need to use special $x$'s to get something. It is a typical idea to approximate $A$ by finite-rank matrices (if you are taking a functional analysis course, you will see more examples like this when you're studying compact operators)... $\endgroup$
    – user58955
    Apr 6, 2020 at 3:18
  • $\begingroup$ and here the easiest thing to do is to take $n\times n$ submatrix, corresponding to taking $x$ with only finitely many nonzero coordinates. Then you will see that UBP helps. $\endgroup$
    – user58955
    Apr 6, 2020 at 3:19
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    $\begingroup$ You can use Cauchy-Schwarz to show that $T_N$ is bounded. $\endgroup$
    – user58955
    Apr 6, 2020 at 8:26

1 Answer 1

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First let's show a simpler version (1-dimensional): If $\sum_i a_i x_i < \infty$ all for $x\in\ell^2$, then $a\in \ell^2$.

You can prove this claim using uniform boundedness principle, or you can just use Riesz Representation Theorem. See this post.

Now, let's go back to your problem. It follows from the claim above that each row of $A$ is in $\ell_2$. Define $T_N$ to be the restriction of $A$ onto the first $N$ rows, that is, $$ T_N x = \left(\sum_j a_{1j}x_j,\sum_j a_{2j}x_j,\dots,\sum_j a_{Nj}x_j,0,0,\dots,\right). $$ We claim that $\|T_N\| < \infty$. Note that $$ \|T_Nx\|_2^2 = \sum_{i=1}^N \left|\sum_j a_{ij}x_j\right|^2 \leq \sum_{i=1}^N\left(\sum_j |a_{ij}|^2 \right)\left(\sum_j |x_j|^2 \right) \leq \|x\|_2^2\cdot \sum_{i=1}^N\sum_{j=1}^\infty |a_{ij}|^2, $$ thus $$ \|T_N\| \leq \left(\sum_{i=1}^N\sum_{j=1}^\infty |a_{ij}|^2\right)^{1/2}. $$ (Note that the infinite sum over $j$ is finite because of the claim at the beginning.)

Now, for each fixed $x$, observe that $\|T_Nx\|_2$ is uniformly bounded by $\|Ax\|_2$ (since $\|T_Nx\|_2$ is just part of the sum for $\|Ax\|_2<\infty$). It follows from the uniform boundedness principle that $\sup_N \|T_N\|<\infty$. Note that $\|Ax\|_2 = \lim_{N\to\infty} \|T_Nx\|_2 \leq (\sup_N \|T_N\|)\|x\|$, which implies that $A$ is bounded and $\|A\| \leq \sup_N \|T_N\|$.

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  • $\begingroup$ In this sentence "(Note that the infinite sum over $j$ is finite because of the claim at the beginning.)" do you mean over $j$ or over $i$? $\endgroup$
    – user593471
    Apr 7, 2020 at 8:20
  • $\begingroup$ I mean over $j$. It's $\sum_{j=1}^\infty$, this is an infinite sum $\endgroup$
    – user58955
    Apr 7, 2020 at 8:23
  • $\begingroup$ I think in the 8th line, in your middle term , the $i$ should be until $N.$ $\endgroup$
    – user593471
    Apr 7, 2020 at 8:23
  • $\begingroup$ Oh, you're right. Let me fix the typo $\endgroup$
    – user58955
    Apr 7, 2020 at 8:24
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    $\begingroup$ @Happy So it's slightly different from your hint. I am taking $T_N$ to be the first $n$ rows of $A$ while in your hint it is the first $n$ columns of $A$. No, I am not taking a special $x$ to say $\sum_j |a_{ij}|^2$ is finite. This is from the 1-dimensional version stated at the beginning of the proof. Since $Ax\in \ell^2$, every coordinate of $Ax$ must be finite. The $i$-th coordinate of $Ax$ is $\sum_j a_{ij}x_j$, and this is finite for any $x\in \ell^2$. The 1-dimensional version implies that the $i$-th row of $A$ is in $\ell^2$, that is, $\sum_j |a_{ij}|^2 < \infty$. $\endgroup$
    – user58955
    Apr 7, 2020 at 9:16

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