If two functions can be shown to agree at an infinite number of points, what additional information would be required to show that these two functions are equivalent?

For example, if two polynomials agree at an infinite number of points, then they are equivalent due to the fundamental theorem. However, $y = 1$ and $\sin(x)$ agree at infinite number of points, but are clearly not the same function.

The context for this question is from the following question. Please feel free to answer it as well.

If $f(x): \mathbb{R} \rightarrow \mathbb{R}$ is a strictly positive continuous function such that $f(c) = c$ for some real $c \ne 0$, and $f(x) = f(x+f(x))$ for all $x$, prove that $f(x) = c$ for all $x$.

  • $\begingroup$ I doubt that you can find a general answer to this. Every function ought to be equivalent to itself, and for some functions there seems to be no short cut to specifying that they be equal at all points - e.g. Dirichlet function $f(x) = 1$ when x is rational; $= 0$ otherwise. $\endgroup$ – Tom Collinge Jun 9 '14 at 22:25
  • $\begingroup$ I assumed he meant two continuous functions. Otherwise you obviously need all points. $\endgroup$ – AnalysisStudent0414 Jun 9 '14 at 22:26
  • $\begingroup$ for f it follows that $f(n.c) = c $ for all integers $n \ge 1$ but I can't get further. $\endgroup$ – Tom Collinge Jun 10 '14 at 12:13
  • $\begingroup$ Before I forget how I did it, here's proof $f(n.c) = c$. True for $n=1$. Suppose true for $n$, i.e $f(n.c) = c$, then $f([n+1].c) = f(n.c + c) = f(n.c + f(n.c)) = f(n.c) = c$ $\endgroup$ – Tom Collinge Jun 10 '14 at 14:03
  • $\begingroup$ And a more general observation that looks like it would lead to the result: for $f(x) = y$, then $f(x + n.y) = y$. Same proof by induction. True for 0; if true for $n$ then $f(x + [n+1].y) = f(x + n.y + y) = f(x+n.y + f(x + n.y)) = f(x + n.y) = y$. $\endgroup$ – Tom Collinge Jun 10 '14 at 14:16

You need them to be the same function on a dense set. It may still be countable, but dense.

The typical example is $\mathbb{Q}\subset \mathbb{R}$, if you have $f,g$ continuous such that $f(q)=g(q)$ $\forall q \in \mathbb{Q}$ then an easy argument shows they are the same function on $\mathbb{R}$ (take $r$ real, there is a $q_n \to r$, apply continuity).

That being said, I'm looking forward to the solution of your problem. I tried to work on it with no result.

  • 1
    $\begingroup$ That's one possibility. On a different approach, if the functions are analytic and equal over an open set ? $\endgroup$ – Tom Collinge Jun 9 '14 at 22:33
  • $\begingroup$ Oh well in complex analysis things are a lot better! I was talking about real functions. $\endgroup$ – AnalysisStudent0414 Jun 9 '14 at 22:35
  • $\begingroup$ There are real analytic functions too en.wikipedia.org/wiki/Analytic_function $\endgroup$ – Tom Collinge Jun 9 '14 at 22:38
  • $\begingroup$ Of course, I always think about those things as restrictions of complex analytic functions, but you're right. Anyway even assuming $f$ real analytic I can't prove the thesis $\endgroup$ – AnalysisStudent0414 Jun 9 '14 at 22:40

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