# How to prove that a function is continuous in functional equation?

I was wondering about different methods or properties to prove that a function in a functional equation is continuous or differentiable. Can somebody give me some examples of such problems or methods, where in the given functional equation you can prove that the function(s) are continuous and a proof of that? I know that if we find the function, it might happen to be continuous, but I'm looking for problems which first prove continuity before finding the solution.

• Is there any particular motivation you have for looking at these kinds of problems? – Omnomnomnom Dec 21 '15 at 14:02
• Well, recently I started solving a lot of functional equations in order to prepare myself for math contests. Since in a lot of problems I see that if $f$ is continuous, I could easily solve the problem, I though it a good idea to see in what ways I could prove such things. But my interest is not only for math contest problems, but also in general. – thefunkyjunky Dec 21 '15 at 14:05

In general, a non-continuous function is non-continuous for a few reasons:

1. It doesn't exist in a certain domain of $x$ values.

2. You get an indeterminate form at $x=a$.

3. You get a division by $0$.

4. The function is a piece-wise function and the limit $\lim_{x\to a}$ does not exist.

5. Asymptotes.

I'm sure there are other reasons why a function would be non-continuous, but I can't think of any more for now.

1. Note any logarithms and square roots, as these are elementary functions that don't exist for negative values. Note that they only exist when the function inside of them is positive (for only real values).

2. If you have taken calculus, then you should be familiar with indeterminate forms: $0^0$, $\infty-\infty$, $\infty^0$, $0^{\infty}$, $0/0$, and $\infty/\infty$, to name most of them. When we have $x=a$ and it produces any of the latter, then we must use calculus methods to determine what $x=a$ comes out to be (or other methods).

3. Pretty obvious, division by $0$. Set all your denominators equal to $0$ to see if this case can occur. The only way this point can exist is if we can cancel with the numerator or do something funky with the function to get it to work.

4. Piece-wise functions, where you get pieces of the function for certain domains of $x$. Check each boundary to see if the function is continuous. For example:

$$f(x)=g(x),x\le0$$$$f(x)=h(x),x>0$$For this, you might want to check whats happening at $x=0$.

1. Asymptotes can occur due to logarithms, divisions by $0$, and simply because of the type of function you are dealing with.

For example, $\tan(x)$, $\frac1x$, $x!$, and others.

Proofs of whether or not these work and how evade me, but it seems pretty logical.

After checking all of that, you want to know if the function is differentiable?

Well, it should be differentiable as long as the function is continuous over a given domain.

For example, we don't say $\frac1x$ is differentiable at $x=0$ but we do say it is differentiable at $x\ne0$.

Got it? Good luck with those competitions!

• So you suppose proving that a function is NOT non-continuous, which would imply that it is continuous? – thefunkyjunky Dec 22 '15 at 5:02
• Yes, I would believe so. If not, it most certainly helps. – Simply Beautiful Art Dec 22 '15 at 14:12