# Validating a Delta Sequence

1. Oct 6, 2012

### Undoubtedly0

The problem statement, all variables and given/known data
Show that $$\delta_n(x) = ne^{-nx} \quad \mathrm{for}\quad x>0$$$$\qquad = 0 \quad \mathrm{for}\quad x<0$$

satisfies $$\lim_{n\longrightarrow\infty}\int_{-\infty}^\infty \delta_n(x)f(x)\mathrm{d}x = f(0)$$

The attempt at a solution
The hint says to replace the upper limit ($\infty$) with $c/n$, where $c$ is "large but finite", and then use the mean value theorem of integral calculus.

I do not understand how this replacement in the hint is allowable. Since $n\longrightarrow\infty$, $c/n\longrightarrow0$, not $\infty$. Even if this is okay, how does it aid using the mean value theorem of integral calculus?

2. Oct 6, 2012

### voko

Are there any restrictions on f(x)?

3. Oct 7, 2012

### Undoubtedly0

It is only stated that $f(x)$ is a well-behaved function. It seems that this would be easier to prove if $\lim_{x\longrightarrow\pm\infty}f(x)$ was at least finite, but no such assumption is made.

4. Oct 7, 2012

### voko

Then assume the integral of f(x) over the entire real line is finite. Given that, you should be estimate the "left-over" as you restrict integration to c/n.

5. Oct 7, 2012

### Undoubtedly0

Why can this be assumed? Not even a function as well-behaved as $f(x)=x^2$ satisfies this.

6. Oct 7, 2012

### voko

$e^{x^{100}}$ is also well behaved, but will it be acceptable in this case?

7. Oct 7, 2012

### Undoubtedly0

Sure, if $f(x) = e^{x^{100}}$, then I do not know how I would answer the problem. Any suggestions?

8. Oct 7, 2012

### voko

As I said, you should select a class of functions that are admissible. Those whose integral over the entire real line is finite obviously fit the bill. Those the product of which with exp(-x) has the same property would work, too.

9. Oct 8, 2012

### Undoubtedly0

Evidently, however, the proof can be made for any well-behaved function $f(x)$. I understand how to prove this for special cases, but I am confused on how to prove this in general.

10. Oct 8, 2012

### voko

$\delta_1 = e^{-x}$ so you must have at least $\int_0^{\infty} f(x)e^{-x} dx < \infty$ for any "well-behaved" function.

11. Oct 20, 2012

### Undoubtedly0

If there exists a Taylor series expansion of f(x) of the form $$f(x)=\sum_{k=0}^\infty f^{(k)}(0)\frac{x^k}{k!}$$

does that imply that $\int_0^{\infty} f(x)e^{-x} dx < \infty$?

12. Oct 20, 2012

### HallsofIvy

Staff Emeritus
You are asked to show this for n a fixed constant. There is nothing said about "n going to infinity" and no reason to worry about that.

13. Oct 20, 2012

### Undoubtedly0

Hi HallsofIvy. I confess I do not understand what you mean. Certainly the following says that n goes to infinity, no?

$$\lim_{n\longrightarrow\infty}\int_{-\infty}^\infty \delta_n(x)f(x)\mathrm{d}x$$