Upper bounding integral Given a probability density function $p(x)$ and a strictly positive function $f(x)$, I am wondering whether it is possible to upper bound $\int_{-\infty}^{\infty} f(x)^2 p(x) dx$ in terms of $\int_{-\infty}^{\infty} f(x) p(x) dx$. I am interested in additive constants, powers, multipliers, etc.
The first idea that jumps into one's mind is Jensen's inequality, but that would provide a lower bound in our case.
 A: Suppose there exists a continuous function $F$ such that
\begin{align}
\int^\infty_{-\infty} f(x)^2 p(x)\ dx \leq F\left(\int^\infty_{-\infty} f(x)p(x)\ dx \right).
\end{align}
Consider $f_N(x) = Nf(Nx)$, then we see that
\begin{align}
\int^\infty_{-\infty} N^2f(Nx)^2 p(x)\ dx = N\int^\infty_{-\infty} f(u)^2p\left(\frac{u}{N}\right)\ du \leq F\left(\int^\infty_{-\infty}f(u)p\left(\frac{u}{N} \right)\ du \right). 
\end{align}
Suppose $p(x)$ is the gaussian and $f \in L^1(\mathbb{R})\cap L^2(\mathbb{R})$, then we see that 
\begin{align}
\lim_{N\rightarrow \infty} F\left(\int^\infty_{-\infty}f(u)p\left( \frac{u}{N}\right)\ du \right) = F\left(\int^\infty_{-\infty} f(u)\ du \right)<\infty
\end{align}
but
\begin{align}
\lim_{N\rightarrow \infty} N\int^\infty_{-\infty} f(u)^2 p\left(\frac{u}{N} \right)\ du = \infty.
\end{align}
Hence it doesn't look too hopeful. 
Edit: It should be noted that there does not exists a constant $C$ such that
\begin{align}
\| f\|_{L^2(dP)} \leq C\|f\|_{L^1(dP)}
\end{align}
is a consequence of my above argument. 
