Is the 4-norm of the orthogonal projection of a vector into a subspace less than or equal to the 4-norm of the vector itself? To simplify the question, suppose $V\subseteq \mathbb{R}^n$ is a subspace and $P$ is the projection matrix. Is it true that $\forall x \in \mathbb{R}^n, \|Px\|_4 \le \|x\|_4$. I guess it is true for any $k$-norm if $k\ge 2$, but I don't know how to prove it.
 A: It's not true. Counterexample:
$$
P=\frac14\pmatrix{1&\sqrt{3}\\ \sqrt{3}&3},\ x=\pmatrix{1\\ 1}.
$$
$P$ is symmetric and $P^2=P$. Hence $P$ is an orthogonal projection. However, $$\|Px\|_4=\frac{(1+\sqrt{3})10^{1/4}}4=1.2146>1.1892=2^{1/4}=\|x\|_4.$$
A: As was concisely pointed out by user1551, the answer is negative, but perhaps the following could shed some more light as to why it is negative, and why $k>2$ is important.
Note first that for every symmetric orthonormal matrix $U$ in $\mathbb{R}^n$, the matrix $P=(U+I)/2$ is a symmetric projection matrix, as can be easily verified. (It is symmetric and $P^2=P$). This, together with the following lemma, allows us to construct orthogonal projections with large $||\cdot ||_k$ norm.
Lemma. Fix a positive integer $m>2$. Let $v_0\in R^{2m}$ denote the vector whose $m$ last coordinates are $1$ and the $m$ first coordinates are zero. Let $u_0\in R^{2m}$ denote the vector whose $m+1$'st coordinate is $\sqrt{m}$ and all the rest are zero. Then there exists an orthonormal symmetric $2m\times 2m$ matrix $U$ such that $Uv_0=u_0$. 
Before I prove the lemma, let me show how it is used to answer the question. Let $P={U+I\over 2}$ with $U$ from the lemma and $I$ the identity in $R^{2m}$. Then:
$$||v_0||_k=m^{1/k},\enspace\hbox{but}\enspace ||Pv_0||_k=||{\sqrt{m}e_{m+1}+v_0\over 2}||_k>{\sqrt{m}\over 2}$$
Here $e_{m+1}$ denotes the corresponding unit vector. Consequently, the norm of the projection as an operator from $R^{2m}$ to $R^{2m}$ is larger than ${1\over 2}m^{1/2-1/k}$, which tends to infinity as $m\to\infty$ when $k>2$.
It remains to prove the lemma.
Proof of lemma. Let $Q:R^{2m}\to R^m$ denote the projection which leaves untouched the last $m$ coordinates and deletes the first $m$ coordinates. There exists an orthonormal matrix $V$ in $R^m$ that maps $Qv_0$ to $Qu_0$, because both have the same Euclidean norm, namely $\sqrt{m}$, and the orthogonal group acts transitively on spheres. Consider the $2m\times 2m$ block matrix:
$$U=\begin{bmatrix}
    0 & V  \\
    V^t &0 \\
  \end{bmatrix}$$
$U$ has the desired properties: it is orthonormal (all columns have norm $1$ and are mutually orthogonal), it is symmetric, and $Uv_0=u_0$. Q.E.D.
