Suppose f is positive and Schwartz function. Fix $N>0$ and $A>0$. Suppose that for any $x \in [-N,N]$, $$A \leq \int_{-N}^{N}f(x-z)dz$$ Then do the inequality $$A \leq C_{r} \frac{1}{N^{1/r}}\int_{-N}^{N}(\int_{-N}^{N}|f(x-z)|^rdx)^{1/r}dz$$ hold for some $C_{r}$ and for any $r>0$?
I solved the case $r \geq 1$. By integration on $[-N,N]$, $$2NA \leq \int_{-N}^{N}\int_{-N}^{N}f(x-z)dxdz$$ Now, just apply Holder inequality. Then we get the desired result.
If $f$ is a monotone function, then the case $0<r<1$ also true but I don't know how to prove general case. Thanks in advance.
EDIT : (Scaling argument)
Put $g(z)=\frac{1}{N}f(\frac{z}{N})$. Then we may assume that $N=1$.
I found the counterexample. (N=1 version)Take f(x)=2n on [1+1/2n,1+1/2n−1]. Then A can be any positive number, but the RHS integral is finite.