Let f(z) be an entire function and f(z) = 0 on a subinterval a < x < b of the real axis. Show that f(z) = 0 for all z in C (set of complex numbers). Conclude that if f and g are entire functions and that they agree on a segment of the real axis, then they must be the same.
How do I answer this question? Thanks for help in advance.
Ok, I'd like to try:
Prove that if on a subset of the real axis , then throughout .
Since is analytic, then we can write for some :
where is analytic and bounded in some neighborhood of , that is for , and is the first non-zero term. Now:
Now recall that we have a dense set of points over which . Therefore, we can find a such that . This means:
This would imply since and but by assumption . This must mean we've reached a contradiction and all so that .
I'll leave it to you to show the function on the same domain implies since can still be written as for some (new) set of .
This may need a little more work . . .
That's a good proof. I knew another way to conclude, which you may like to read:
Originally Posted by shawsend
So we have chosen and we have written for any . We want to show that for all It is well-known that (derivation term by term, and then ). This derivative is a complex derivative, and since is real, it coincides with the derivative of restricted to the real line (same expression as a limit). But is 0 near , hence , and subsequently .
You used the word "dense", it is a bit improper, and it could be " is an accumulation point of ", or " is not an isolated point in ", to underline the tight relation with what you said about non-isolated subsets where is zero. The proof of the fact you quoted is then the very same.
I wish to attempt a summary. Please someone correct if necessary. I struggle with this.
First note the Taylor series for a function is unique. That is, anyway we can determine the Taylor series for the function centered at a point will be the Taylor series. Now, is real on the real axis in the set . This mean we can calculate the Taylor series of by calculating the Taylor series for the real function over . but is zero in that interval which means all the derivatives are zero as well (it's flat). This means all the constants in the real Taylor expansion are zero. But since Taylor series are unique, this Taylor series must be the Taylor series for as well, thus throughout .