A direct proof involving a positive summability kernel

[psie]

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.

 

[fresh_42]

$n\cdot f(s)$ 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 $x=0.$

 

[martinbn]

The mean value theorem gives you that for each $n$ the integral is equal to $f(c_n)$ for some $-\frac1{2n}\le c_n\le \frac1{2n}$.

 

[psie]

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...

 

[fresh_42]

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!

 

[pasmith]

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, <br /> \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). and by continuity of f at zero <br /> \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).

 

[psie]

by continuity of f at zero <br /> \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).

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.

 

[mathwonk]

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).