Show δn = (sin nx) / (pi x) is a delta distribution

[ognik]

Given:
Assume that f (x) is continuous at x = 0 and vanishes as x→±∞.
Hint. Replace x by y/n and take lim n→∞ before integrating.

I apply the hint and get $ \frac{1}{\pi}\int_{-\infty}^{\infty} \,{\lim_{{n}\to{\infty}}}{\frac{sin y}{y}} \,dy $
For the limit, as n -> ${\infty}$ , y -> 0. By L'Hospitals rule, $ {\lim_{{n}\to{\infty}}}{\frac{sin y}{y}} =>\lim_{{y}\to{0}} cos y$ = 1
I am pretty sure this is not right, but can't see any other way to approach this? (sin y)/y is not integrable...

 

[Opalg]

Given:
Assume that f (x) is continuous at x = 0 and vanishes as x→±∞.
Hint. Replace x by y/n and take lim n→∞ before integrating.

I apply the hint and get $ \frac{1}{\pi}\int_{-\infty}^{\infty} \,{\lim_{{n}\to{\infty}}}{\frac{sin y}{y}} \,dy $
For the limit, as n -> ${\infty}$ , y -> 0. By L'Hospitals rule, $ {\lim_{{n}\to{\infty}}}{\frac{sin y}{y}} =>\lim_{{y}\to{0}} cos y$ = 1
I am pretty sure this is not right, but can't see any other way to approach this? (sin y)/y is not integrable...

Just a few comments about this.

First, please post the question in the body of the text, not just in the title (see MHB's Rule #10).

Second, $$\lim_{n\to\infty}\frac{\sin y}{y}$$ is $\dfrac{\sin y}{y}$, not $1$ (the limit is taken with respect to $n$, not $y$, so it is the limit of a constant as far as $n$ is concerned).

Third, $\dfrac{\sin y}{y}$ is not integrable in the sense of Lebesgue, but it is integrable as an improper Riemann integral, and $$\int_{-\infty}^{\infty}\frac{\sin y}{y}dy = \pi.$$

 

[chisigma]

The definition of delta distribution as $\displaystyle \delta (x) = \lim_{n \rightarrow \infty} \delta_{n} (x)$ where $\displaystyle \delta_{n} (x) = \frac{\sin n x} {\pi\ x}$ is, in my opinion, a little questionable because none of the $\delta_{n} (*)$ is a distribution function because they aren't non negative...

... better the definition, among other, with $\displaystyle \delta_{n} (x) = \frac{n}{\sqrt{\pi}}\ e^{- n^{2}\ x^{2}}$ ...

Kind regards

$\chi$ $\sigma$

 

[ognik]

Thanks for the answers - and for the protocol tips, being new I appreciate that.
I can see clearly what you said, much appreciated, would like some clarity on a couple of points please.

Because I replaced x with y/n, y=nx, y is a function of (x,n), both are variables? So how can I treat y as constant in the limit?
Before I posted, I did come across the solution of $\pi$ you suggest - but that hasn't been covered in the text I am following, so I wanted to see if there was a more 'from 1st principles' approach? Thanks.

 

[ognik]

Hi again, I have been working through the text again and have found more questions than answers ...

If δ(y) is a dirac distribution, then I should be able to show that δ(0) = $\infty$ and δ(y)=0, x $\ne$ 0,
$\lim_{{y}\to{\infty}} (n/\pi)\frac{sin y}{y}=0$ (because $sin y \le 1 $ and iff n is finite)
$\lim_{{y}\to{0}} (n/\pi)\frac{sin y}{y}=n\pi$ instead of $\infty$, so this wants n infinite and contradicts the above?
I am missing something I suspect?

Also, I should be able to show that $\lim_{{n}\to{\infty}}\int_{-\infty}^{\infty}f(x) \,d \frac{sin nx}{\pi x}=f(0)$
I can see from the graph of $ \frac{sin nx}{\pi x}$ that δ(y) is an adequate dirac distribution, but ...
Previously we arrived at $\lim_{{n}\to{\infty}}\int_{-\infty}^{\infty} \, \frac{sin nx}{\pi x}=1$ - is there way to get f(0) without first proving δ(y) is a dirac distribution?

 

[chisigma]

Hi again, I have been working through the text again and have found more questions than answers ...

If δ(y) is a dirac distribution, then I should be able to show that δ(0) = $\infty$ and δ(y)=0, x $\ne$ 0,
$\lim_{{y}\to{\infty}} (n/\pi)\frac{sin y}{y}=0$ (because $sin y \le 1 $ and iff n is finite)
$\lim_{{y}\to{0}} (n/\pi)\frac{sin y}{y}=n\pi$ instead of $\infty$, so this wants n infinite and contradicts the above?
I am missing something I suspect?

Also, I should be able to show that $\lim_{{n}\to{\infty}}\int_{-\infty}^{\infty}f(x) \,d \frac{sin nx}{\pi x}=f(0)$
I can see from the graph of $ \frac{sin nx}{\pi x}$ that δ(y) is an adequate dirac distribution, but ...
Previously we arrived at $\lim_{{n}\to{\infty}}\int_{-\infty}^{\infty} \, \frac{sin nx}{\pi x}=1$ - is there way to get f(0) without first proving δ(y) is a dirac distribution?

If You define the Dirac distribution as $\displaystyle \delta (x) = \lim_{n \rightarrow \infty} \delta_{n} (x)$ where $\displaystyle \delta_{n} (x) = \frac{\sin n x}{\pi\ x}$, You have also to verify that $\displaystyle \int_{- \infty}^{+ \infty} \delta (x)\ d x = 1$...

... that is true because for any $\delta_{n} (*)$ is ...

$\displaystyle \frac{1}{\pi} \int_{- \infty}^{+ \infty} \frac{\sin n x}{x}\ d x = \frac{1}{\pi}\ \int_{- \infty}^{+ \infty} \frac{\sin \xi}{\xi}\ d \xi = 1\ (1)$

It seems to be 'all right' but one has to take into account that a function f(x) is a distribution function not only if $\displaystyle \int_{- \infty}^{ + \infty} f(x)\ d x = 1$ but also if $\displaystyle f(x) \ge 0,\ \forall x$ and that isn't true for the $\delta_{n} (x)$...

Kind regards

$\chi$ $\sigma$

 

[ognik]

If I understand chisigma's reply, the problem does say "Assume that f (x) is continuous at x = 0 and vanishes as x→±∞. Sorry I didn't include that, I was looking only at δ(x)"

So to summarise, the queries I still have are:
1. Because I replace x with y/n, y=nx, y is surely a function of (x,n), both are variables? So how can I treat y as constant in the limit?
2. If δ(y) is a dirac distribution, then I should be able to show that δ(0) = ∞ and δ(y)=0, x ≠ 0
But all I can see is
$ \lim_{{y}\to{0}} \frac{siny}{y} =n\pi$ instead of ∞, so this wants n infinite?
3. and similarly, $ \lim_{{y}\to{\infty}} (\frac{n}{\pi})\frac{siny}{y}=0 $ instead of ∞, so this wants n finite which contradicts the above?

I'd appreciate some understanding on the above, thanks

 

[Opalg]

If I understand chisigma's reply, the problem does say "Assume that f (x) is continuous at x = 0 and vanishes as x→±∞. Sorry I didn't include that, I was looking only at δ(x)"

So to summarise, the queries I still have are:
1. Because I replace x with y/n, y=nx, y is surely a function of (x,n), both are variables? So how can I treat y as constant in the limit?
2. If δ(y) is a dirac distribution, then I should be able to show that δ(0) = ∞ and δ(y)=0, x ≠ 0
But all I can see is
$ \lim_{{y}\to{0}} \frac{siny}{y} =n\pi$ instead of ∞, so this wants n infinite?
3. and similarly, $ \lim_{{y}\to{\infty}} (\frac{n}{\pi})\frac{siny}{y}=0 $ instead of ∞, so this wants n finite which contradicts the above?

I'd appreciate some understanding on the above, thanks

There are some things that need to be sorted out in the statement of this problem. First, you include the condition "Assume that $f (x)$ is continuous at $x = 0$ and vanishes as $x\to\pm\infty$", but there is no further mention of the function $f$ in the remainder of the question. Second, you don't seem to be making any distinction between the functions $\delta_n$ and the distribution $\delta$.

The only way that I can make sense of the problem is to assume that you are being asked to show that the sequence $\{\delta_n\}$ converges to $\delta$ (in the sense of convergence of distributions) as $n\to\infty$.

Remember that the Dirac distribution $\delta$ is not a function, and statements like $\delta(y)=0$ or $\delta(0)=\infty$ make no mathematical sense (though they are intuitively useful). In fact, the distribution $\delta$ acts on functions, by the formula $\delta(f) = f(0).$ A function such as $\delta_n$ can also act as a distribution, by the formula $$\delta_n(f) = \int_{-\infty}^\infty \delta_n(x)f(x)\,dx.$$

To show that $\{\delta_n\} \to \delta$ as $n\to\infty$, you need to show that $\delta_n(f) \to\delta(f)$ for every function $f$ specified in the statement of the question. In other words, you must show that $$\lim_{n\to\infty} \int_{-\infty}^\infty\frac{\sin(nx)}{\pi x}f(x)\,dx = f(0).$$ The first step is to make the substitution $x = y/n$, getting $$\lim_{n\to\infty} \int_{-\infty}^\infty\frac{\sin(nx)}{\pi x}f(x)\,dx = \lim_{n\to\infty} \int_{-\infty}^\infty\frac{\sin y}{\pi y}f\bigl(\tfrac yn\bigr)\,dy.$$ The question now says "take lim $n\to\infty$ before integrating". This is mathematically dubious, but since you are told that you are allowed to do it, you can go ahead and find that $$\lim_{n\to\infty} \int_{-\infty}^\infty\frac{\sin(nx)}{\pi x}f(x)\,dx = \int_{-\infty}^\infty\lim_{n\to\infty} \frac{\sin y}{\pi y}f\bigl(\tfrac yn\bigr)\,dy = \int_{-\infty}^\infty \frac{\sin y}{\pi y}f(0)\,dy.$$ From there, you should be able to complete the integral as discussed earlier in this thread, getting the answer $f(0)$ as required.

 

[ognik]

Thanks for that reply - the book I am following is mathematical methods for physicists, I gather physicists sometime shade mathematical formality ...

So having previously arrived at $ \lim_{{n}\to{\infty}}\int_{-\infty}^{\infty} \, sin (nx)/\pi x = 1 $ , I thought I had already satisfied what the question wanted. f(0) is effectively a constant, so $ \lim_{{n}\to{\infty}}\int_{-\infty}^{\infty}f(0) \,sin (nx)/\pi x = f(0)*1 = f(0) $

But I was left with the 3 queries itemized above and would like clarity on those for my deeper understanding please? Thanks...