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.


1 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 $t\to+\infty$).

  • $\begingroup$ Thanks, that makes sense. $\endgroup$ Aug 27, 2023 at 2:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.