Let be a continuous function in and satisfies . Prove that is identically zero .
Well, I'm having a go at , which seems more sensible. To start with, putting for shows that , so we can concentrate on the interval .
So put for to derive (1).
Putting gives , or (2).
It is clear from (2) that .
Putting in (1) and (2) gives , so now .
Put in (1) and (2) now shows that .
Now direct substitutions, the identity and previous results show that .
So far we know that
.
You can see how to proceed. We can prove by induction that for all positive integers , whenever .
The function is continuous on and vanishes on . Moreover is dense in (anyone else want to prove this?)
It follows that on and therefore on and therefore on .