Continuity of a linear functional. Let $X=c_{00}$ with the norm $\vert \vert . \vert \vert _{p}, 1<p \leq \infty.$ For $r \geq 0$ consider the linear functional $f_{r}$ on $X$ defined by 
$$f_{r}(x)=\sum _{j=1} ^ \infty \frac {x(j)}{j^r},$$
when $x\in X$ then $f_r$ is continuous iff $r>1-1/p$ and then 
$$\vert \vert f_r \vert \vert _{p}=(\sum _{j=1}^\infty \frac{1}{j^{rq}})^{\frac{1}{q}}$$
Please help. Is there a need for Holder Inequality while proving continuity part? 
 A: We have $r>1-1/p,$ and so $qr>q-q/p=1$. Therefore, if $\|x\|_p\le 1,$ then 
$a).\ |f_{r}(x)|=\left | \sum _{j=1} ^ \infty \frac {x(j)}{j^r}  \right |\le \|x\|_p\cdot \left ( \sum_{j=1}^{\infty}\left ( \frac{1}{j^{r}} \right )^{q} \right )^{1/q}\le \left ( \sum_{j=1}^{\infty}\left ( \frac{1}{j^{r}} \right )^{q} \right )^{1/q}$ so $f_r$, which converges because $qr>1$. Hence, $f_r$ is bounded  hence continuous.
$b).$ Now suppose $r\le1-1/p$ and $f_r$ is continuous. Extend $f_r$ to a continuous linear functional, which we still call $f_r$, on all of $l_p.$ Since $0<r<1$, we may choose $0<s<1$ such that $r+s<1.$ If we take $x_n=\underbrace{(1,(1/2)^{s},(1/3)^{s},\cdots,(1/n)^{s}}_{\text{n}},0,0\cdots, )$ then $x_n$ converges in $l_p$ because $p/s>1$. But $|f_{r}(x_n)|=\left | \sum _{j=1} ^n \frac {1}{j^{(r+s)}}\right |$, which diverges, which contradicts the continuity of $f_r$.
$c).$ Assume $0<p<\infty.$  By part $a).$, we have $\|f_r \|\le \left ( \sum_{j=1}^{\infty}\left ( \frac{1}{j^{r}} \right )^{q} \right )^{1/q}.$ But $l_p$ is reflexive so $f_r$ attains it norm. This proves that  $\|f_r \|= \left ( \sum_{j=1}^{\infty}\left ( \frac{1}{j^{r}} \right )^{q} \right )^{1/q}.$
On the other hand, $p=\infty$, we have$\|f_r \|\le \sum_{j=1}^{\infty}\frac{1}{j^{r}},$ so if we take $x=(1,1,\cdots, ),$ then $x\in l_{\infty}$ and $f_r(x)=\sum_{j=1}^{\infty}\frac{1}{j^{r}},$ as desired. 
