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Consider the two linear spaces: $l^{2} = \left\{x = (x_1, x_2, . . . ) : \sum_{k=1}^{\infty} |x_k|^{2} < \infty\right\}$ with norm $||x||_{2} = (\sum_{k=1}^{\infty} |x_k|^{2})^{\frac{1}{2}}$, and $l^{\infty} = \left\{x = (x_1, x_2, . . . ) : \sup_{k} |x_{k}| < \infty\right\}$ with norm $||x||_{\infty} = \sup_{k} |x_{k}|$

Fix $a \in l^{2}$. Define the function T on $l^{2}$ by $(Tx)_{j} = \sum_{k=1}^{j} a_{k}x_{k}$

(a) Show that T is a continuous linear operator from $l^{2}$ to $l^{\infty}$. Find the norm of T. Justify your answer.

(b) Is T a continuous linear operator from $l^{2}$ to $l^{2}$? If so, find its norm. If not, prove that.

I tried several ways to prove the continuous part, but couldn't prove it based on the definition of continuity. Hope someone can help me with this problem.

Edit: Below is my effort to prove part (a) and (b) thanks to Javier's help:

(a) To show $T_{a}(x)\in l^{\infty}$, we use CS inequality and get:

$(\sum_{k=1}^{\infty} |a_{k}x_{k}|)^2\leq (\sum_{k=1}^{\infty} |a_{k}|^2) (\sum_{k=1}^{\infty} |x_{k}|^2) < \infty$, since $a, x\in l^{2}$.

Thus $\sum_{k=1}^{\infty} |a_{k}x_{k}|< \infty$, which implies $\sup_{k} |a_{k}x_{k}| <\infty$. Since $a$ is fixed, the last inequality implies $\sup_{k} |x_{k}|<\infty$

Now, we will show that $T_{a}(x)$ is bounded. WLOG, we can assume $||x||_{2}= (\sum_{k=1}^{\infty} |x_{k}|^{2})^{\frac{1}{2}}=1$. By using Javier's hint below, we use CS inequality and get:

For the first case:

$(\sum_{k=1}^{N} |a_{k}x_{k}|)^2\leq (\sum_{k=1}^{N} |a_{k}|^2) (\sum_{k=1}^{N} |x_{k}|^2) \leq (\sum_{k=1}^{N} |a_{k}|^2) $, since $a, x\in l^{2}$ and $\sum_{k=1}^{\infty} |x_{k}|^{2}=1$.

Thus, $\sum_{k=1}^{N} |a_{k}x_{k}| \leq (\sum_{k=1}^{N} |a_{k}|^2)^{\frac{1}{2}} < \infty$. Done for this case.

For the infinity case, we use the same trick, and will end up with $\sum_{k=1}^{\infty} |a_{k}x_{k}| \leq (\sum_{k=1}^{\infty} |a_{k}|^2)^{\frac{1}{2}} < \infty$, since $a\in l^{2}$.

(b) First, since $a\in l^{2}$, by using CS inequality, $T_{a}(x)\in l^{2}$. For the boundedness part, WLOG, assume $||x||_{2}=1$. Let $m=\sum_{j=1}^{\infty} |a_{j}|^{4}$, and $n=\sum_{j=1}^{\infty} |x_{j}|^{4} \leq 1$. Note that since $x,a\in l^{2}$, $m$ and $n <\infty$. Now, by Holder inequality, we have:

$||T_{a}(x)||_{2} = (\sum_{j=1}^{\infty} |a_{j}|^{2}|x_{j}|^{2})^{\frac{1}{2}}\leq m^{\frac{1}{2}}n^{\frac{1}{2}}\leq m^{\frac{1}{2}}$, which is bounded since $a\in l^{2}$. Thus T is a linear continuous operator from $l^{2}$ to $l^{2}$

Martin Sleziak
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ghjk
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  • What is your input? – Bombyx mori Sep 27 '14 at 05:11
  • I haven't had a clear approach yet:( – ghjk Sep 27 '14 at 13:37
  • It is unusual to just post a (homework) question here without indicating your own thoughts and difficulties. So: What are your thoughts, what have you tried, where are you stuck? Do you know the relevant definitions (if not, look them up), do you know the Cauchy-Schwarz inequality? – PhoemueX Sep 27 '14 at 21:32
  • @PhoemueX: thank you for your feedback! I will definitely add more into it. On this particular problem, I was stuck right at the beginning (not knowing where to start:P) – ghjk Sep 28 '14 at 04:01
  • I am pretty sure that $T$ is not bounded from $\ell^2 \to \ell^2$. To see this, try to calculate the norm of $T(a)$. – PhoemueX Sep 28 '14 at 07:35
  • I tried above, but I got bounded. I'm trying to come up with a particular a and x to give a counter-example, but not successful yet:P You have any examples? Can you see where my proof went wrong, since I haven't found a mistake:( – ghjk Sep 28 '14 at 14:42
  • can anyone please help verify if my solutions above are good or not? I spent quite an effort to type them up:) – ghjk Sep 29 '14 at 22:00
  • Since it has been a while, anybody can verify my solution above for part (a)? And is there any proof to show $T$ is not bounded from $l^2\rightarrow l^{\infty}$? – ghjk Oct 25 '14 at 03:59

2 Answers2

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b) I might have misunderstood the problem but I don't think that $T$ necessarily even maps $\ell_2$ into $\ell_2$.

Just take $a=x=(1,1/2,1/3,\dots)$, i.e., $a_k=x_k=\frac1k$.

Clearly $a,x\in\ell_2$.

We have $(Tx)_n=\sum_{k=1}^n \frac1{k^2} = 1+\frac1{2^2}+\dots+\frac1{n^2}$. We have $\lim\limits_{n\to\infty} (Tx)_n = \frac{\pi^2}6$. If $Tx$ belonged to $\ell_2$, then this limit would be zero. (Necessary condition for convergence of a series.)

Martin Sleziak
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  • Great counter-example, Martin! Many thanks for your help! I have a quick question though: the contradiction to "convergence criteria" is because of the fact that $l^2$ is a Hilbert space in this case. I wonder whether a counter-example exists if we replace $l^2$ by some other non-Hilbert and infinite-dimensional linear spaces?? – ghjk Dec 03 '14 at 18:13
  • I'd say that in works because of $x\in\ell_2$ $\Rightarrow$ $\sum x_n^2<+\infty$ $\Rightarrow$ $\lim\limits_{n\to\infty} x_n =0$. This is not directly related to the fact that $\ell_2$ is Hilbert space. What I'm using there is that $\ell_2\subseteq c_0$. (And also that it contains the sequence used in counterexample.) – Martin Sleziak Dec 03 '14 at 19:12
  • What is $c_0$ in this case?Btw, did you think of my 2nd question? – ghjk Dec 03 '14 at 21:57
  • @user177196 Sorry for not explaining the notation, $c_0$ is the space of all sequences convergent to zero (a.k.a. null sequences). It is usually taken with sup-norm, but here I am using this notation merely as the notation for the set of all such sequences. (I think this notation is pretty standard in functional analysis, it is mention in Wikipedia article on sequence spaces.) – Martin Sleziak Dec 04 '14 at 05:23
  • @user177196 So that we avoid too long discussion in the comment, we can try to continue the discussion in chat. (At least if you are somewhat familiar with using SE chat.) But I have to admit that I am probably not able to add much more to what I have already written there. – Martin Sleziak Dec 04 '14 at 05:38
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Hint for the first part:

a) First use Hölder's inequality to show that $T_a(x) \in l_\infty$. Then it's sufficient to show that $T$ is bounded for it to be continous.

$$\|T_a\| = \sup\left\{\|T_a(x)\|_\infty:\|x\|=1\right\}=\sup\left\{\left\lvert \sum_{j=1}^Na_jx_j \right\rvert\,\,\text{or}\,\, \left\lvert\sum_{j=1}^\infty a_jx_j\right\lvert:\,x=1\right\}$$

Can you find the supremum in each of the two possible cases?

b) At the moment I don't have an answer for this. If I come up with something I'll add it.

Martin Sleziak
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hjhjhj57
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  • I have 2 questions though: (1) Why bounded implies continuity? My guess is you're using the definition "epsilon - delta" of continuity in this case, is that right? (2) Where do you get the first equality of $||T_{a}||$ (i.e, where does $x=1$ come from), and shouldn't the 2nd expression only has $j$ from $1$ to $\infty$, by definition of norm $||x||_{\infty}$ above. – ghjk Sep 27 '14 at 14:03
  • Many thanks for your hint! To show $T_{a}(x)\in l^{\infty}$, by CS inequality, $(\sum_{k=1}^{j} |a_{k}x_{k}|)^2\leq \sum_{k=1}^{j} |a_k|^2 \sum_{k=1}^{j} |x_{k}|^2 < \infty$, since $a, x\in l^{2}$. Since $\sum_{k=1}^{j} |a_{k}x_{k}| \geq sup |a_{k}x_{k}|$ and $a_k$ is fixed, $sup |x_{k}| < \infty$. For the bounded part, assume $x=1$ means $x_{j}=1$, the first case < $N\infty=\infty$ by CS inequality. The 2nd case < $\infty$ by the trick: $(\sum_{j=1}^{\infty} |a_j|)^{2} = (sum_{j=1}^{\infty} |a_j|^{2} - 2$(an expression) $< \infty$ since $a\in l^{2}$. – ghjk Sep 27 '14 at 14:12
  • (1) The equivalence is an important fact in functional analysis, see: http://en.wikipedia.org/wiki/Bounded_operator#Equivalence_of_boundedness_and_continuity (2) You should also check the equivalent definitions of the norm of an operator and try to prove them: http://en.wikipedia.org/wiki/Operator_norm#Equivalent_definitions No, it's possible that the finite sum is bigger than the infinite one, can you think of an specific example? – hjhjhj57 Sep 27 '14 at 18:28
  • To show $T_a(x)\in l_\infty$ you should use the Hölder inequality, it's easy and you don't need to consider two cases. – hjhjhj57 Sep 27 '14 at 18:33
  • I really didn't consider two cases for $T_{a}(x)\in l_{\infty}$, did I? Btw, is my proof above correct? And how is the finite sum bigger than the infinite one when these two sums consider the same terms, as you used the same notations for two sums above? If they are different terms, I think of Riemann-Zeta sum for p=2? – ghjk Sep 27 '14 at 18:39
  • How about $a=(a_1,a_2,\cdots, a_n,-1, -1/2, -1/4, -1/8,...)$ $x=(a_1,a_2,\cdots ,a_n,,1, 1/2, 1/4, 1/8,...)$? And about your proof two things: The first part is right, but your sums should be infinite, don't they? And for the ' bounded part' is wrong, what I meant by $x=1$ is $||x||=1$, otherwise we wouldn't need to take the suprermum. – hjhjhj57 Sep 27 '14 at 19:04
  • Thank you for your example, Javier! However, all the terms in the supremum are in absolute value:) I'd like to correct my proof for the bounded part (and yes, the sums in the first part is infinite): since $||x||{2}=1$ and $x\in l^{2}$, $(\sum{j=1}^{\infty} |x_j|^{2})^{\frac{1}{2}}=1$. Now, using CS inequality, both the first and 2nd case < $\infty$ since $(\sum_{j=1}^{\infty} |a_{j}x_{j}|)^{2}\leq (\sum_{j=1}^{\infty} |a_{j}|^2 )(\sum_{j=1}^{\infty} |x_{j}|^2) < \infty$. Done with part (a) now? – ghjk Sep 28 '14 at 04:12
  • Btw, I think we can set $||x||{2} = any positive constant$, since we still get the upper bound for $||T{a}||$ by using CS inequality. Now, my question is: why did you set $||x||_{2}=1$? Are we allowed to do this W.L.O.G? – ghjk Sep 28 '14 at 04:16
  • Check the links I already provided. As suggested above, you should either edit or make another question (I don't know which would the right option be) where you write down everything you've already done along with all the questions that arise in the process. – hjhjhj57 Sep 28 '14 at 04:41
  • I'd like to give part (b) a try now. First, since $a\in l^{2}$, by CS inequality, $T_{a}(x)\in l^{2}$. Now, for the boundedness part, let $m=\sum_{k=1}^{\infty} |a_{j}|^2 <\infty$, and $n=\sum_{k=1}^{\infty} |x_{j}|^2 <\infty$, since $x\in l^{2}$. Now, by AM-GM and Holder inequality, we have: $||T_{a}(x)||_{2}\leq \frac{1}{2}(m+n)<\infty$. Thus T is a linear continuous operator, but I don't know how to find a norm:( – ghjk Sep 28 '14 at 04:53
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    I added my solutions to both parts in the original post now:) (costed me a bunch of time, but I was glad I got it done) – ghjk Sep 28 '14 at 05:40
  • could you verify if my solution for part (b) is correct? I haven't been able to find a counter-example, if there is one:( – ghjk Sep 29 '14 at 00:59
  • I don't know, but be careful. I think (not sure) that you're using the Hölder inequality in some places and not CS, as you say (the difference is wether you take the absolute value of the whole sum (CS) or the sum of the absolute vaules (Hölder)). – hjhjhj57 Sep 29 '14 at 02:23
  • I used the Holder for absolute values across part (a) and (b). For part (b), since all the terms are squared, I think it doesn't matter though. Did you get the solution for part (b) though? – ghjk Sep 29 '14 at 04:24