$a \in \mathbb{R}^N, fixed$ and $T: \mathbb{R}^N \longrightarrow \mathbb{R}, continous$
$T(x) = \langle a,x \rangle = \sum_{i=1}^Na_ix_i$
Norm in $\mathbb{R}^N$ is $| \cdot | _p, p \in [1,\infty]$ and norm in $\mathbb{R}$ is $| \cdot |$.
Show that the operator norm $||T|| = |a| _{\frac{p}{p-1}},$ for $1 < p < \infty$.
I have already been able to show that $||T|| \leq |a| _{\frac{p}{p-1}}$ using Hölder's inequality but how can I show that $||T|| = \sup_{x \in \mathbb{R}^N\ \backslash \{0\}} \frac{|T(x)|}{|x| _p} = |a| _{\frac{p}{p-1}}$? Thanks.
Take $x_i = \mathrm{sgn}({a_i}) \; |a_i|^{1/(p-1)}$ for each $i$, where $\mathrm{sgn}$ is the sign function.
$$T(x) = \sum_{i=1}^N a_i x_i = \sum_{i=1}^N \underbrace{a_i \; \mathrm{sgn}({a_i})}_{|a_i|} \; |a_i|^{1/(p-1)} = \sum_{i=1}^N |a_i|^{p/(p-1)}$$
$$|| x ||_p = \left(\sum_{i=1}^N |x_i|^p \right)^{1/p} = \left(\sum_{i=1}^N |a_i|^{p/(p-1)} \right)^{1/p}$$
$$|| a ||_{p/(p-1)} = \left(\sum_{i=1}^N a_i^{p/(p-1)} \right)^{(p-1)/p} = \frac{|T(x)|}{||x||_p}$$
Note that $x \ne 0$, so we're done.