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

Question

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.

Answer 1

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 $t\to+\infty$).