The question is as follows:

Prove the linear boundedness of $f(x) = \sum_{k=1}^{+\infty} \frac{x_k}{\sqrt{k}}$, for $x=(x_k)_k \in \ell_p$ for $1 \leq p < 2$. And find its norm.

$\textbf{some effort:}$

For to show it is bounded, we have $||fx||_{\ell_p}^{p} = \sum_{l=1}^{+\infty}\sum_{k=1}^{\ell} |\frac{x_k}{\sqrt{k}}|^p$

$\hspace{7.6cm} \leq \sum \sum \frac{|x_k|}{\sqrt{k}}^p$

$\hspace{7.6cm} = \sum \mid \frac{1}{\sqrt{k}} \mid^p \sum |x_k|^p$.

This means that $||fx|| \leq (\sum \mid \frac{1}{\sqrt{k}} \mid^p)^{\frac{1}{p}} (\sum |x_k|^p)^{\frac{1}{p}}$. And then $||f|| \leq \sum_{k=1}^{+\infty} \frac{1}{\sqrt{k}}$.

Please correct me if I am so far of being correct.

Now we need to prove that $||f|| \geq \sum_{k=1}^{+\infty} \frac{1}{\sqrt{k}}$?

Can you please help me to find its norm?


  • $\begingroup$ That sum is convergent for all $x\in \ell_p$ only when $p < 2$. In those cases, it defines a bounded linear functional. $\endgroup$ Dec 9 '17 at 19:35
  • $\begingroup$ @DanielFischer Thanks! Oh sorry, you are right! I will correct it now. What about my calculation? $\endgroup$
    – Nikita
    Dec 9 '17 at 19:39
  • 1
    $\begingroup$ $p$ is still less than $2$ in your post. what is the correct range? $\endgroup$
    – daulomb
    Dec 9 '17 at 20:12
  • $\begingroup$ @daulomb Thanks! Yes it is less than 2. Later I corrected it. But your answer is correct. $\endgroup$
    – Nikita
    Dec 9 '17 at 20:16

Hölder's inequality $\Rightarrow$ $$|f(x)|\leq \displaystyle\bigg(\sum_{k=1}^{\infty}\frac{1}{k^{q/2}}\bigg)^{1/q}\displaystyle\bigg(\sum_{k=1}^{\infty} |x_k|^p\bigg)^{1/p}\leq \displaystyle\bigg(\sum_{k=1}^{\infty}\frac{1}{k^{q/2}}\bigg)^{1/q}\|x\|_{\ell_p}$$ $\Longrightarrow$ $$\|f\|\leq \bigg(\sum_{k=1}^{\infty}\frac{1}{k^{q/2}}\bigg)^{1/q}.$$

But here $q$ must be greater than $2$ in order for the sum to exist. For the other direction think about a specific element of $\ell_p$. Tha is, take an $x\in\ell_p$ and use the fact that $|f(x)|\leq \|f\|\|x\|_{\ell_p}.$

  • $\begingroup$ Many Thanks! Can we choose $x$ such that $x^{\frac{1}{1-p}} = (\frac{1}{k^{\frac{1}{2(p-1)}}})$? Then we have $||x^{\frac{1}{p-1}}|| = (\sum_{k=1}^{+\infty} \frac{1}{k^{\frac{p}{2(p-1)}}})^{\frac{1}{p}}$ . And then we have $||x||_{ell_p} = (\sum_{k=1}^{+\infty} \frac{1}{k^{\frac{p}{2(p-1)}}})^{\frac{1-p}{p}} = \bigg(\sum_{k=1}^{\infty}\frac{1}{k^{q/2}}\bigg)^{1/q}$? $\endgroup$
    – Nikita
    Dec 9 '17 at 20:10

Are you sure about your claim? If i choose $x_k=\frac{1}{\sqrt k}$, then $x \in \ell _p, \forall p >2$ but $\sum_{k=1}^{k=+\infty}\frac{1}{k}=+\infty$, so $f$ can't be bounded.

Also in your try you wrote $\|fx\|_{\ell_p}^p$, but $f:\ell_p \rightarrow \mathbb R$ so $f(x) \in \mathbb R$ is a real number and the norm of $fx$ is just the norm of $\mathbb R$.

If you want to show that it is bounded for all $x \in l_p, p<2$ then it can be done using the Hölder's inequality. Note that if $p<2$ then $y:y_k=\frac{1}{\sqrt k} \rightarrow y\in \ell_{p'}$ so we have: $$\sum_{k=1}^{+\infty}\left| \frac{x_k}{\sqrt k}\right|=\|xy\|_{\ell_1}\le\|x\|_{\ell_p}\|y\|_{\ell_{p'}}.$$ Now we have:

$$|f(x)|\le\|xy\|_1 \le k \cdot \|x\|_p.$$ That shows that $f$ is bounded.

Bonus section, if $p=1 \rightarrow\|f\|=\|y\|_{\infty}=1$, in fact take $x_1=1,x_k=0, k\neq 1$, then $$|f(x)|=1=1\cdot \|x\|_1=\|y\|_{\infty}\|x\|_1.$$

  • $\begingroup$ Many thanks! Yes you are right. That was my mistake and later I corrected it. Thanks for your answer and for the bonus section! $\endgroup$
    – Nikita
    Dec 9 '17 at 20:14
  • $\begingroup$ You can take $x=(x_k)\in \ell_p$ with $x_k=\frac{1}{k^{\frac{q-1}{2}}}$. But you will need $p(q-1)>2$ where $p$ and $q$ are exponents in Hölder's inequality. $\endgroup$
    – daulomb
    Dec 9 '17 at 20:28
  • $\begingroup$ This choice will give $\|f\|=\bigg(\sum_{k=1}^{\infty}\frac{1}{k^{q/2}}\bigg)^{1/q}=\|x\|_{\ell_q}.$. $\endgroup$
    – daulomb
    Dec 9 '17 at 20:34
  • $\begingroup$ @daulomb Thanks! But I am so sorry, I have problem with $|f(x)|\leq \|f\|\|x\|_{\ell_p}.$. Because the point of using this is to find $x \in \ell_p$ such that if we put it in $|f(x)|\leq \|f\|\|x\|_{\ell_p}.$ then by letting it comes to the left side, then we should get $(\sum \frac{1}{k^{\frac{q}{2}}})^{\frac{1}{q}} |f(x)| \leq ||f||.$ But in this case for to show that $(\sum \frac{1}{k^{\frac{q}{2}}})^{\frac{1}{q}} \leq ||f||.$ we need to also have $|f(x)| = 1$? Am I right? $\endgroup$
    – Nikita
    Dec 9 '17 at 20:40
  • $\begingroup$ You just compute $f(x)$ and $\|x\|_{\ell_p}$ and use the fact I suggested in the answer you will see that $\|f\|\geq\|x\|_{\ell_q}$. Combining this with the other direction you get the equality. But dont forget $1/p+1/q=1$. $\endgroup$
    – daulomb
    Dec 9 '17 at 20:46

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