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

[alphaone]

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.

 

[HallsofIvy]

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).

 

[alphaone]

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.

 

[HallsofIvy]

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.

 

[alphaone]

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)?

 

[alphaone]

If my question is inclear please let me know, because I really would like to know the answer of this question.

 

[HallsofIvy]

Without more information on f(m,n) you can only conclude that A= f(m,m)