Can delta function equality hold for m not equal to n?

In summary, the conversation discusses an equation of the form delta(m-n)*A=delta(m-n)*B and the conditions for which A=B can be concluded. It is mentioned that this argument came up in a paper of a Nobel laureate and the question is raised about the validity of deducing A=lim(m->n)f(m,n) from the equation. The expert summarizer concludes that without more information on f(m,n), it can only be deduced that A= f(m,m) or f(m,n).
  • #1
alphaone
46
0
Hi
I am not a mathematician so my question might be silly.
I really came across it in physics but I think it is purely mathematical:
I came across an equation of the form:
delta(m-n)*A= delta(m-n)*B
my question is now for what cases can I conclude A=B?
Does this only hold for m=n, or can I also conclude A=B for m->n ,i.e. m=n+epsilon for sufficiently small epsilon. I am not sure about this as the naive definition of the delta function(infinite at 0; 0 everywhere else and integrates to 1 when 0 is in the integration range) would lead me to conclude that only when m=n I could conclude A=B. However the delta function is the limit of a set of function which do not vanish for for m=n+epsilon so maybe their is an argument why for epsilon small enough I should be able to conclude A=B. Another reason why I think this should be the case is that this argument came up in a paper of a nobel laureate and his argument crucially depended on the possibility to conclude A=B even when m not equal n but only m=n+epsilon. So if anybody could give me a rigorous argument why this should be possible I would be thankful to hear about it.
 
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  • #2
What are A and B? Numbers or functions of x?

If numbers, then yes, A= B. If functions of x, then you can only conclude that A(m-n)= B(m-n).
 
  • #3
Thanks for the reply.
Sorry the way I phrased the question was really stupid. Obviously if A and B are const then A=B, but for the case I am interested we have A=const and B=f(m,n). Now my quesion should have been:
If we have delta(m-n)*A=delta(m-n)*f(m,n) then in a famous paper a nobel laureate concludes A=lim(m->n)f(m,n) from which he deduces the validity that A=f(m,n) for m=n+epsilon. It is explicitly stated there that he does not assume lim(m->n)f(m,n)=f(m,m) but that he only needs lim(m->n)f(m,n).
So I was wondering why A=lim(m->n)f(m,n) if he does not assume lim(m->n)f(m,n)=f(m,m). However there must be a way as it is a celebrated paper and without this possibility the main prrof would not work.
 
  • #4
alphaone said:
Thanks for the reply.
Sorry the way I phrased the question was really stupid. Obviously if A and B are const then A=B, but for the case I am interested we have A=const and B=f(m,n). Now my quesion should have been:
If we have delta(m-n)*A=delta(m-n)*f(m,n) then in a famous paper a nobel laureate concludes A=lim(m->n)f(m,n) from which he deduces the validity that A=f(m,n) for m=n+epsilon. It is explicitly stated there that he does not assume lim(m->n)f(m,n)=f(m,m) but that he only needs lim(m->n)f(m,n).
So I was wondering why A=lim(m->n)f(m,n) if he does not assume lim(m->n)f(m,n)=f(m,m). However there must be a way as it is a celebrated paper and without this possibility the main prrof would not work.
The value of f(n,n) (strictly speaking, since m is going to n you want f(n,n), not f(m,m)) is completely irrelevant to lim(m->n)f(m,n).

The definition of "lim(x->a) f(x)= L" is "For any [itex]\epsilon>0[/itex], there exist [itex]\delta>0[/itex] such that if [itex]0< |x-a|< \delta[/itex] then [itex]|f(x)-L|< \epsilon[/itex]." Notice the "0< "? What happens at x=a doesn't matter. In general, you can change f(a) to be anything at all (or even leave it undefined) without changing lim(x->a) f(x).

Of course, if f is continuous at a, then lim(x->a)f(x) must equal f(a) but apparently that is not being assumed here.
 
  • #5
Thanks for the reply. I see what you are saying but now my quetion is:
In the case of delta(m-n)*A=delta(m-n)*f(m,n) is it valid to deduce A=lim(m->n)f(m,n) or is it only valid to deduce A=f(m,m)?
 
  • #6
If my question is inclear please let me know, because I really would like to know the answer of this question.
 
  • #7
Without more information on f(m,n) you can only conclude that A= f(m,m)
 

1. What is a delta function?

A delta function, also known as a Dirac delta function, is a mathematical function that is defined as zero everywhere except at the origin, where it is infinite. It is often used in physics and engineering to represent a point source or impulse.

2. What is the purpose of a delta function?

The purpose of a delta function is to act as a mathematical tool for representing a point source or impulse. It allows us to simplify calculations and accurately model physical phenomena such as point charges or forces.

3. How is a delta function different from a regular function?

A regular function assigns a value to each point in its domain, while a delta function only has a value at the origin. Additionally, a delta function has a value of zero everywhere except at the origin, where it is infinite.

4. What are some applications of the delta function?

The delta function has many applications in physics and engineering, including in the study of electromagnetic fields, quantum mechanics, and signal processing. It is also used in the theory of distributions, which is a mathematical framework for generalizing functions.

5. Is the delta function a real physical entity?

No, the delta function is a mathematical construct and does not physically exist. It is a useful tool for representing physical phenomena, but it does not have a physical form or substance.

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