Consider the space $\mathbb{E}:=\mathbb{R}^n$ to be equipped with the norm
$$||x||_p:=\Bigg(\sum_{k=1}^n |x_k|^p\Bigg)^{\frac{1}{p}},\,\,\,\,x=(x_1,...,x_n)^T\in\mathbb{E}$$
where $p>1$. Consider the space $\mathbb{E}^*$ and let $f\in\mathbb{E}^*$ be given by
$$\langle f,x\rangle:=\sum_{k=1}^n x_ky_k$$
Compute
$$||f||:=\sup_{||x||_p\le 1}\langle f,x\rangle$$
So far I did the following
BY uisng Holder's inequality we have
$$\langle f,x\rangle=\sum_{k=1}^n x_ky_k\le||x||_p ||y||_q$$
Therefore
$$||f||=\sup_{||x||_p\le 1}\langle f,x\rangle\le\sup_{||x||_p\le 1}||x||_p ||y||_q\le ||y||_q$$
I am not quite sure whether I am right so far, also I am stuck with the reverse way. Any help would be highly appreciated. Thanks in advance.
Now, if we take $x_i=sig(y_i)|y_i|^{q-1}$, where $sig(y_i)$ is the sign of the real number $y_i$, we have that
$\lVert x \rVert_p=\left(\sum_{i=1}^n |y_i|^{qp-p}\right)^{\frac1p} = \left(\sum_{i=1}^n |y_i|^{q}\right)^{\frac1p}$. Now:
$$\langle f,x \rangle=\sum_{i=1}^n |y_i|^q = \left(\sum_{i=1}^n |y_i|^q \right)^{\frac 1p+\frac 1q}=\left(\sum_{i=1}^n |y_i|^q \right)^{\frac 1p}\left(\sum_{i=1}^n |y_i|^q \right)^{\frac 1q}=\lVert x\rVert_p \lVert y\rVert_q $$
So $\langle f,\frac{x}{\lVert x \rVert} \rangle=\lVert y\rVert_q$.