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 $q=4$?



Simultaneous near-best approximation with respect to two norms
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.


