# Convolution and Smoothness Conditions

Suppose $f(x),g(x)\in L_1(\mathbb{R})$, with both $|f(x)| \leq 1$, $|g(x)| \leq 1$ and $|f(x)| \rightarrow 0$, $|g(x)| \rightarrow 0$ for $|x| \rightarrow \infty$. Given that we have two other functions $f_1,g_1$ such that for large enough $|x|$, $|f(x)|\geq |f_1(x)|$, and $|g(x)| \geq |g_1(x)|$, what can I then say about the convolution

$$f*g(x) = \int f(y)g(x-y)dy ?$$

I suspect that I am about to bound from below $|f*g|$ by something like $|f_1(x)g_1(x)|$ or $|f_1(x)| + |g_1(x)|$ but I cannot seem prove it.

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Pointwise? No chance. There is nothing to prevent $(f*g)(x)$ from being zero at any particular point, with $f_1,g_1\ne 0$ at that point. –  user53153 Feb 15 '13 at 2:52
Do you have a specific example in mind? I should have perhaps added that $f_1$ and $g_1$ are non - trivial in the sense that they aren't the $0$ functions, but proper decaying functions –  user61038 Feb 15 '13 at 2:57
I should also say I'm interested in its asymptotic limit behaviour. –  user61038 Feb 15 '13 at 3:10
Consider the following possibility: $f(x)=f_1(x)=\exp(-x^2)$, $g(x)=\exp(-x^2)\sin(\phi(x))$, $g_1(x)=|g(x)|$. Here $\phi(x)=x^2$ or perhaps $\phi(x)=\exp(\exp(x))$ - the point is that $g$ has a lot of rapid oscillation which will make $f*g$ extremely small due to cancellation. Since $g_1$ does not change sign, no such cancellation happens in $f_1*g_1$, or $f_1+g_1$, etc - these are functions with Gaussian behavior at infinity. –  user53153 Feb 15 '13 at 3:29
Alright ok, thankyou –  user61038 Feb 15 '13 at 4:05