almost-summability kernal with jump discontinoualities to zero

Question

A summability kernel $\phi_n$ is a sequence of continuous real valued function with the requirement of 3 properties. Convolution of a function $f$ with with such a summability kernal coverges uniformly to $f$. That is to say $$\lim_{n\to\infty} f*\phi_n = f$$

I am looking at a couple of "almost summability kernels". Which can described more or less as as unit area rectangles. For example:

$$g_{n}(x)=\begin{cases} \dfrac{1}{2n} & -n<x\le n\\ 0 & otherwise \end{cases}$$

Except for it not being continuous, it has all the properties that are a required of a summability kernal.

Integrates to 1:$$ \int_{-\infty}^{\infty}g_{n}(x)\,dx=\int_{-**\infty}^{-n^{-}}0dx+\int_{-n^{+}}^{n^{-}}\frac{1}{2n}dx+\int_{n^{+}}^{\infty}0dx=0+\left[\frac{t}{2n}\right]_{-n}^{n}=\frac{2n}{2n}=1$$

Integral of Absolute value bounded: Always non negative so $$\int_{-\infty}^{\infty}|g_{n}(x)|\,dx=\int_{-\infty}^{\infty}g_{n}(x)\,dx=1$$

Limit of sequence tail integral zero : $$\lim_{n\to\infty}\int_{|x|\ge\delta}|g_{n}(x)|\,dx=\lim_{n\to\infty}\int_{|x|\le\delta}g_{n}(x)\,dx=\lim_{n\to\infty}\int_{\delta}^{\infty}g_{n}(x)dx+\int_{-\infty}^{-\delta}g_{n}(x)dx$$

By the symmetry of the function about x=0

this is equal to $$\lim_{n\to\infty}2\int_{\delta}^{\infty}g_{n}(x)dx=\lim_{n\to\infty}\begin{cases} \dfrac{n-\delta}{n} & \delta<n\\ 0 & n<\delta \end{cases}=0$$

So $$g_n$$ this would be a summability kernel except that it is not continuous.

Does it still have the property that convolving a function with its terms converges to that function?

Answer 1

The error is in the "Limit of sequence tail integral zero" part. Intuitively, the integral of the tail is 1 minus the integral in $[-\delta,\delta]$ since the integral is always 1, and we know that this integral goes to zero since the value in $[-\delta,\delta]$ monotonically goes to zero (thus use the Monotone Convergence Theorem), meaning the integral of the tail goes to 1.

Or you can see the error in your proof is that you correctly calculated

$$\lim_{n\to\infty}2\int_{\delta}^{\infty}g_{n}(x)dx=\lim_{n\to\infty}\begin{cases} \dfrac{n-\delta}{n} & \delta<n\\ 0 & n<\delta \end{cases}$$

but finished this wrong. Since $n\rightarrow\infty$ we can drop the constant zero case for any finite $\delta$, so it's just $\lim_{n\rightarrow\infty} \frac{n-\delta}{n} = 1$.

Thus the limit of the tail integral is non-zero (1), meaning that this kernel cannot be an Almost Summability Kernel.

Answer 2

No, I do not believe it is, No uniform convergence at least Consider as a counter-example a simple function: $$f(x)=x$$

Then $$\left|g_{n}\ast f(t)-f(t)\right|=\left|\dfrac{1}{2n}\int_{t-n}^{n-t}\tau d\tau-t\right|$$ $$ = \left|\dfrac{1}{2n}\left[\dfrac{t^{2}}{2}\right]_{t-n}^{n-t}-t\right|$$ $$ = \left|\dfrac{(n-t)^{2}-(t-n)^{2}}{4n}-t\right|$$ $$ = t$$

For Uniform convergence, it is required that $\forall\epsilon\in\mathbb{R}$ $\exists N\in\mathbb{R}$ such that $\forall n>N$,$\forall t\in\mathbb{R}$ we have $\left|g_{n}\ast f(t)-f(t)\right|<\epsilon$

But as shown for the convolution of your kernel and the $f(x)=x$, any bound ($\epsilon$) would depend on $t$