A direct proof involving a positive summability kernel

  • #1
psie
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Homework Statement
Prove directly that if $$K_n(s)=\begin{cases}n &\text{if }|s|<1/(2n)\\ 0 &\text{if }|s|>1/(2n),\end{cases}$$ and ##f## is a continuous function at the origin, then $$\lim_{n\to\infty}\int_\mathbb{R}K_n(s)f(s)ds=f(0).$$
Relevant Equations
Positive summability kernels, see e.g. Wikipedia.
This is an exercise from Fourier Analysis and its Applications by Vretblad.

I know the integral over ##\mathbb R## reduces to $$\int_{-1/(2n)}^{1/(2n)} nf(s)ds.$$ But I don't know where to go from here. There is a theorem in the book which states that this limit exists and equals ##f(0)##, but I'm apparently not supposed to use the theorem. Appreciate any help.
 
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  • #2
##n\cdot f(s)## is continuous and therefore integrable, say ##\int n\cdot f(s)\,ds = F(x).## Hence, if ##F(x)## is continuous at the origin, we get
$$
\int_{-1/(2n)}^{1/(2n)} nf(s)\,ds = F(1/(2n))-F(-1/(2n)) \stackrel{n\to \infty }{\longrightarrow }0
$$
So all it has to be shown is, that ##F(x)## is continuous at ##x=0## or at least that ##\displaystyle{\lim_{n \to \infty}}F(1/(2n))=0.## Maybe, we need to split the integral at ##x=0.##
 
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  • #3
The mean value theorem gives you that for each ##n## the integral is equal to ##f(c_n)## for some ##-\frac1{2n}\le c_n\le \frac1{2n}##.
 
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  • #4
Hmm, thanks for the replies. But ##f## is only continuous at ##0##. Can we then say that it has an antiderivative? I'm afraid we can't use the mean value theorem either, which requires continuity in an interval. Maybe there's something missing in the exercise...
 
  • #5
Continuous at ##0## means in an open neighborhood of ##0.## Just choose ##n## large enough. The MVT provides a sequence of mean values ##c_n## that can be trapped in ##\left[-1/(2n)\, , \,1/(2n)\right].## In general, one has to be careful with the MVT when the intermediate value depends on boundaries!
 
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  • #6
psie said:
Homework Statement: Prove directly that if $$K_n(s)=\begin{cases}n &\text{if }|s|<1/(2n)\\ 0 &\text{if }|s|>1/(2n),\end{cases}$$ and ##f## is a continuous function at the origin, then $$\lim_{n\to\infty}\int_\mathbb{R}K_n(s)f(s)ds=f(0).$$
Relevant Equations: Positive summability kernels, see e.g. Wikipedia.

This is an exercise from Fourier Analysis and its Applications by Vretblad.

I know the integral over ##\mathbb R## reduces to $$\int_{-1/(2n)}^{1/(2n)} nf(s)ds.$$ But I don't know where to go from here. There is a theorem in the book which states that this limit exists and equals ##f(0)##, but I'm apparently not supposed to use the theorem. Appreciate any help.

By definition, [tex]
\tfrac1n \inf_{|x| \leq \tfrac 1{2n}} f(x) \leq \int_{-1/2n}^{1/2n} f(x)\,dx \leq \tfrac1n \sup_{|x| \leq \tfrac 1{2n}} f(x).[/tex] and by continuity of [itex]f[/itex] at zero [tex]
\lim_{n \to \infty} \inf_{|x| \leq \tfrac 1{2n}} f(x) = f(0) = \lim_{n \to \infty} \sup_{|x| \leq \tfrac 1{2n}} f(x).[/tex]
 
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  • #7
pasmith said:
by continuity of [itex]f[/itex] at zero [tex]
\lim_{n \to \infty} \inf_{|x| \leq \tfrac 1{2n}} f(x) = f(0) = \lim_{n \to \infty} \sup_{|x| \leq \tfrac 1{2n}} f(x).[/tex]
Could you explain why, by continuity of ##f##, \begin{align} \lim _{n\to \infty }\inf _{x\in \left[-\frac{1}{n}{,}\frac{1}{n}\right]}f\left(x\right)&=f(0), \\ \lim _{n\to \infty }\sup _{x\in \left[-\frac{1}{n}{,}\frac{1}{n}\right]}f\left(x\right)&=f(0).\end{align} I'd be very grateful for your reply.
 
  • #8
if Jn is a sequence of intervals shrinking down to 0, i.e. all containing zero and with diameters approaching zero, and if for each n, Kn is the smallest interval containing all values of f on Jn, then continuity of f at 0, implies that the sequence Kn shrinks down to f(0). note that the endpoints of Kn are exactly the inf and sup of the values of f on Jn.

i.e. as x gets closer to zero, the values f(x) must get closer to f(0). so as n-->infinity, the values of f(x) for -1/n ≤ x ≤ 1/n, must approach f(0).
 
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FAQ: A direct proof involving a positive summability kernel

What is a positive summability kernel?

A positive summability kernel is a sequence or function used in mathematical analysis and harmonic analysis to approximate identities or functions. It typically satisfies certain properties such as positivity, normalization, and tends to zero away from a central point. Examples include Fejér kernels and Dirichlet kernels used in Fourier analysis.

What is a direct proof in the context of summability kernels?

A direct proof in the context of summability kernels involves demonstrating a mathematical statement or theorem by straightforward logical reasoning based on established axioms, definitions, and previously proven results. It does not rely on indirect arguments such as contradiction or contraposition.

Why are positive summability kernels important in analysis?

Positive summability kernels are important because they help in approximating functions and understanding convergence properties. They are used in Fourier analysis to study the convergence of Fourier series and in signal processing to reconstruct signals from their components. Their properties ensure that approximations converge in a controlled manner.

Can you provide an example of a direct proof involving a positive summability kernel?

An example of a direct proof might involve showing that the convolution of a function with a positive summability kernel converges to the function itself as the kernel becomes more concentrated. For instance, one might prove that the convolution of a continuous function with the Fejér kernel converges uniformly to the function.

What are the key properties of a positive summability kernel used in direct proofs?

The key properties include positivity (the kernel values are non-negative), normalization (the integral or sum of the kernel over its domain is 1), and localization (the kernel values tend to zero as one moves away from the central point). These properties ensure that the kernel can approximate the identity function and help in proving convergence results.

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