Let $D=\{\phi_i\}_{i=1}^N$ be an orthonormal set in $L_2[0,1]$, $u(x)\in \operatorname{Span}\{D\}$. Let $\mu(x)$ be a density function on $[0,1]$. Under what condition can we say $\int_{0}^1 u(x)^4 d\mu(x)\leq c\left(\int_0^1 u(x)^2 d\mu(x)\right)^2$ for some constant $c$?
An answer with imposing additional condition on $\mu(x)$ is: $L^{p}$ inequality with a lower bound on measure
I am wondering if we impose some conditions on $\{\phi_i\}_{i=1}^N$, can we still get the same type of argument?
Thanks!
If $N$ is finite, then it is necessary and sufficient that the basis functions be in $L^4$.
Your inequality implies that $u$ is in $L^4$, and clearly each basis function is in the span of $D$.
On the other hand, the span of $D$ is finite-dimensional, and in finite dimension, all norms are equivalent. Thus, in the span of $D$, there is some constant $c$ such that $\|u\|_{L^4}\leq c\|u\|_{L^2}$.