Simultaneous near-best approximation with respect to two norms Suppose that $M$ is a closed infinite dimensional subspace of $L_4(0,1)$ which is also a closed subspace of $L_1(0,1)$. Hence $M$ is isomorphic to $\ell_2$ as a subspace of $L_p(0,1)$ for $1\leq p\leq 4$. Note also that $L_4(0,1)\subset L_1(0,1)$ and $\|f\|_1\leq \|f\|_4$ for each $f\in L_4(0,1)$. 
Is it possible to find a constant $C$ such that for each $f\in L_4(0,1)$ we can find $g\in M$ satisfying 
$\|f-g\|_q \leq C\inf_{h\in M}\|f-h\|_q$ for both $q=1$ and $q=4$?
 A: It seems that the answer is yes.
Denote by $M_q$ the space $M$ equipped with the $L_q$-norm; by the closednedd condition, it is a Banach space for both $q=1$ and $q=4$. The identical mapping $M_4\to M_1$ is bounded; by the bounded inverse theorem, so is its inverse. Thus there exists $\mu$ such that $\|g\|_4\leq \mu\|g\|_1$ for all $g\in M$; surely $\mu\geq 1$. We claim that $C=2+2\mu$ works.
To show this, take any $f\in L_4(0,1)$ and denote $\alpha_q=\rho_q(f,M)=\inf_{h\in M}\|f-h\|_q$. Choose $g_1,g_4\in M$ such that $\|f-g_q\|_q\leq 2\alpha_q$ and set $g=(\alpha_4g_1+\alpha_1g_4)/(\alpha_1+\alpha_4)$. We claim that $g$ fits the goal.
Firstly, notice that 
$$
  \|g_1-g_4\|_1\leq \|g_1-f\|_1+\|f-g_4\|_1\leq 2\alpha_1+2\alpha_4,
$$
so $\|g-g_1\|_1=\frac{\alpha_1}{\alpha_1+\alpha_4}\|g_4-g_1\|_1\leq 2\alpha_1$. On the other hand, since $\|g_1-g_4\|_4\leq \mu\|g_1-g_4\|_1\leq 2\mu(\alpha_1+\alpha_4)$ we have $\|g-g_4\|_4\leq 2\mu\alpha_4$. Therefore,
$$
  \|f-g\|_1\leq \|f-g_1\|_1+\|g_1-g\|_1\leq 4\alpha_1\leq C\alpha_1
$$
and
$$
  \|f-g\|_4\leq \|f-g_4\|_4+\|g_4-g\|_4\leq (2+2\mu)\alpha_4=C\alpha_4,
$$
as required.
