# A comparison between a function and its convolution

Assume that $$f$$ is a $$L^p$$ integrable function for $$1\le p\le P_0$$, with $$P_0$$ a positive constant. L is a smooth compactly supported function. Define $$L_\epsilon(x) = 1/\epsilon^n L(x/\epsilon)$$. Is it possible to give an estimate of the measure of the set S, where $$S=\{x\in R^n: L_\epsilon*f(x)\ge\lambda f(x)\}$$, in term of $$\epsilon$$ and $$\lambda$$?

• Would $(L_{\epsilon}*f)(x)$ be better notation than $L_{\epsilon}*f(x)$? – Acccumulation Dec 3 '18 at 23:51
• When there is no confusion, I would prefer the second one... – Fei Wang Dec 4 '18 at 22:37