A delta function conjecture: almost any function can be a delta kernel

Question

I have been thinking about delta kernels, and I think I have come up with a surprising result:

If $f\colon\Bbb{R}\to\Bbb{R}$ is such that $\int_{-\infty}^\infty f(x)\,dx=L$ is finite and nonzero, then $g_k(x)=\frac kLf(kx)$ is a delta kernel.

Note that by "delta kernel", I mean a sequence of integrable functions $g_k(x)$ such that for any absolutely integrable function $h\colon\Bbb{R}\to\Bbb{R}$ that is continuous at $0$, $\lim_{k\to\infty}\int_{-\infty}^\infty g_k(x)h(x)\,dx=h(0)$.

I find this an interesting problem to think about, so I thought I'd post it here. I'll post the proof as an answer if no one else does (unless I've missed some extra hypotheses, in which case someone will point it out, I'm sure).

Answer 1

(Note: $\int f$ will be used throughout in place of $\int_{-\infty}^\infty f(x)\,dx$.) Suppose $f:\Bbb{R}\to\Bbb{R}$ is integrable on $\Bbb{R}$, with $\int f=L\ne0$. By the definition of the integral $\int_{-\infty}^\infty$, we know that the one-sided limits $\lim_{y\to\infty}\int_0^y f=L_+$ and $\lim_{y\to-\infty}\int_y^0 f=L_-$ exist and $\int f=L_++L_-=L$. Let $F$ be the antiderivative of $f$, i.e. $F(y)=\int_0^yf(x)\,dx$.

Then $\lim_{x\to-\infty}F(x)=-K_-$ and $\lim_{x\to\infty}F(x)=K_+$. Taking $g_k(x)=\frac kLf(kx)$ we have

$$\int_0^yg_k(x)\,dx=\int_0^{ky}\frac 1Lf(x')\,dx'=\frac 1LF(ky).$$

By the limit, given an $\varepsilon>0$ we can find an $M$ such that $|F(x)-K_+|<|L|\varepsilon$ for all $x\ge M$, and so for any $y>0$ we can choose $k=\lceil M/y\rceil$ so that $ky\ge M$ and hence

$$\left|\int_0^yg_k(x)\,dx-\frac{K_+}L\right|=\frac1{|L|}\left|F(ky)-K_+\right|<\varepsilon,$$

so the function $G_k(y)=\int_0^yg_k(x)\,dx$ is pointwise convergent to $G_\infty(y)=\frac{K_+}L$ for $y>0$. By similar considerations, $G_\infty(y)=-\frac{K_-}L$ for $y<0$ (and $G_\infty(0)=0$, for what it's worth). By this point, it should be clear that $g_k$ is a delta kernel (noting that the difference between $\frac{K_+}L$ and $-\frac{K_-}L$ is $1$).

Note in particular that nowhere did we assume that $f$ was nonnegative, nor even absolutely convergent.

Answer 2

There is already a theorem which resembles your result:

Suppose $d$ is a non-negative function with the property:$$\int_{-\infty}^{+\infty }d(s)\,ds=1$$Then the sequence $d_k(t):=k\cdot d(kt)$ is a Dirac sequence.

A Dirac sequence has the properties:

1. $d_k\geq0,\forall k$
2. $\int_{-\infty}^{+\infty }d_k(s)\,ds=1,\forall k$
3. $\forall r>0$ and $\forall \varepsilon>0,$ there exists $N\in\mathbb N$ such that $\forall k>N$ we have $\int_{\mathbb{R}\setminus[-r,r]}d_k(s)\,ds<\varepsilon$