$f(x) = \sum_{k=1}^{+\infty} \frac{x_k}{\sqrt{k}}$ is bounded for $x=(x_k)_k \in \ell_p$ and find its norm. 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?
Thanks!
 A: 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 $|f(x)|\leq \|f\|\|x\|_{\ell_p}.$
A: 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 
$$|f(x)|=1=1\cdot \|x\|_1=\|y\|_{\infty}\|x\|_1.$$
