Lp norm of Hadamard matrix What is the Lp norm of the $N$-dimensional Hadamard matrix $H = ((-1)^{i \cdot j})_{i,j}$ for $p > 2$? I know that $\|H\|_1 = N$, $\|H\|_2 = \sqrt{N}$, $\|H\|_\infty = N$ but I can't figure out what it is for other values of $p$. Can we at least give a good upper-bound on it?
Here I consider the induced norm: $\|H\|_p = \max_{x : \|x\|_p = 1} \|Hx\|_p$
(a previous version of this question incorrectly said that $\|H\|_\infty = 1$)
 A: Important Edit: As J.J Green pointed out, the OP contains an incorrectly stated value for $\|H\|_{\infty}$, which I copied without checking below. Interpolating between $(1,\infty)$ using the corrected version would give the trivial bound $\|H\|_p \leq N$. You regain the sharp bound by interpolating instead between $(1,2)$ and $(2,\infty)$.

I assume $L^p$ norm means the operator norm on $\mathbb{R}^N$ with the $\ell_p$ norm.
Then by Riesz-Thorin-Stein interpolation, you have
$$ \|H\|_p \leq \|H\|_1^{1/p} \|H\|_\infty^{1-1/p} = \sqrt[p]{N} $$
By testing on the vector $(1,0,0,\ldots,0) \mapsto (-1,1,-1,1,\ldots)$ you have
$$ \|H\|_p \geq \sqrt[p]{N} $$
and hence $\sqrt[p]{N}$ is the value.
A: I'm reluctant to contradict Professor Wong, but I think the accepted answer is incorrect since it takes $\left\|H\right\|_\infty$ as 1, whereas it is in fact $N$.  But the same argument applies, with Riesz in this case being
$$
\left\| H \right\|_p \leq \left\| H \right\|^{2/p}_2 \left\| H \right\|^{1-2/p}_\infty
= N^{1/p} N^{1 - 2/p} = N^{1 - 1/p} = N^{1/q}
$$
with $q$ the conjugate index to $p$. This inequality is realised by a vector with equal non-zero entries and so gives the equality.
One can confirm this using software such as Matlab or Octave which implement Higham's approximation for $L^p$ norms of matrices for non-trivial $p$, and that gives a lower bound:
octave:1> format long
octave:2> norm(hadamard(2), 4)
ans =  1.68179283050743
octave:3> norm(hadamard(4), 4)
ans =  2.82842712474619
octave:4> norm(hadamard(8), 4)
ans =  4.75682846001088

