# Does convolution of a compactly supported function with Gaussian need to have fraction of the $L_1$ mass in the original interval?

Let $$f \in L_1(\mathbb{R})$$ be such that $$\operatorname{supp} f \subset [0,1]$$, and let $$K$$ be the gaussian kernel $$K(t) := \frac{1}{\sigma \sqrt{2 \pi}} \exp(-t^2/2\sigma^2)$$, with some small $$\sigma < 1/8$$.

Is it true that for some universal constant $$c > 0$$, we have $$\|\mathbf{1}_{[0,1]} \cdot (K \ast f)\|_1 \geq c \|K \ast f\|_1$$? I.e. is it necessary that at least constant fraction of the $$L_1$$ mass of $$K\ast f$$ stays inside the interval $$[0,1]$$?

This is rather easy to show when $$f$$ is non-negative (say, if $$\|f\|_1 = 1$$, we have $$\|K \ast f\|_1 = \|f\|_1 = 1$$, and $$\|(1 - \mathbf{1}_{[0,1]}) \cdot (K \ast f)\|_1 \leq (\Phi(0) + (1 - \Phi(1/\sigma)) \leq 0.6$$ where $$\Phi$$ is CDF of Gaussian distribution), but in general case some magical cancellations are a priori possible.

Discreete case looks false

Experimentally, the statement seems to be false in the discrete case: let us consider functions on integers $$\mathbb{Z}$$ and a kernel corresponding to $$n$$ step random walk $$K_n := L\ast L \ast L \ast \cdots \ast L$$, where $$L(t) = \frac{1}{3}$$ for $$t \in \{-1, 0, 1\}$$ (and $$0$$ otherwise). I'm chosing non-zero probability for staying in the same place, to avoid period $$2$$ in the random walk.

Let us fix some $$\sigma < 1/8$$. Now if we consider a function $$f : \mathbb{Z} \to \mathbb{R}$$ such that $$\mathrm{supp} f \subset \{0, \ldots n\} =: D$$, we would like to say that for any such function $$\|\mathbf{1}_D \cdot (K_{\sigma^2 n^2} \ast f)\| \geq C \|K_{\sigma^2 n^2} \ast f\|_1$$ for some universal constant $$C$$ that do not depend on $$f$$ nor $$n$$.

But the matrix $$M_{i,j} = K_{\sigma^2 n^2}(i-j)$$ for $$i,j \in \{0, \ldots n\}$$ is invertible (although terribly conditioned), so we can just chose some sparse $$\tilde{f} : \{0, \ldots n \} \to \mathbb{R}$$, which we intent to be $$K_{\sigma^2 n^2} \ast f$$ restricted to $$\{0, \ldots, n\}$$, say $$\tilde{f}(\lfloor n/10\rfloor) = 1$$, and $$0$$ otherwise. We can now solve for $$f := M^{-1} \tilde{f}$$, and extend $$f$$ to all integers (by setting $$f(t) = 0$$ for $$t \not \in D$$).

After solving it numerically it seems that $$f$$ this provides a counterexample. We have $$[K_{\sigma^2 n^2} \ast f]|_{D} = \tilde{f}$$ by construction, hence $$\|\mathbf{1}_D \cdot (K_{\sigma^2 n^2} \ast f)\|_1 = 1$$, while outside of the region $$D$$, the $$\|(1 - \mathbf{1}_D)\cdot (K_{\sigma^2 n^2} \ast f)\|_1$$ is growing with $$n$$ quite drastically.

This exact construction does not seem to lead to a counterexample for the continuous case --- the functions $$f$$ we get by solving $$M^{-1} \tilde{f}$$ do not seem to converge to anything.

No by the usual duality nonsense. Let's say $$\sigma=1$$ (it doesn't matter). The convolution at $$x$$ is $$e^{-x^2/2}$$ times the integral of $$g(t)=f(t)e^{-t^2/2}$$ against $$e^{tx}$$. If the estimate $$\left|\int_2^3(f*K)e^{x^2/2}\,dx\right|\le C\int_0^1|f*K|e^{x^2/2}\,dx$$ were possible, then there would exist an $$L^\infty$$ function $$h$$ on $$[0,1]$$ such that $$\int_{0}^1 h(x)e^{tx}\,dx=\int_2^3 e^{tx}\,dx$$ for almost all $$t\in[0,1]$$, which is clearly impossible (both sides are analytic functions in $$t$$ and the RHS grows faster than the LHS can afford as $$t\to+\infty$$).