# Derivates and continuity.

I was reading a book about fourier analysis, and asked myself a question i wasnt able to answer.

Suppose we have a function f: $\mathbb{R} \to \mathbb{R}$ and that it is piecewise continuous, that is, it has a finite number of discontinuities which are simple discontinuities, that is, $f(x+)$ and $f(x-)$ exist, with the possibilities that $f(x+) \not= f(x-)$ or $f(x+)=f(x-) \not= f(x)$.

If we have a point $x_0$ such that $f(x_{0}+) = f(x_{0}-)$ and the left hand side derivative aswell as the right hand side derivative exist at this point, can we say that $f(x_{0}+) = f(x_{0}-) = f(x_0)$ ?

• If the derivative from the left exists at $x_0$, then $f$ is continuous from the left at $x_0$. Feb 20, 2014 at 20:37
• So you mean that if the derivative exist from the left side, than $f(x-)$ exist and is equal to $f(x)$?
– user117449
Feb 20, 2014 at 20:49
• Yes (though you already assumed $f(x_0-)$ exists). So then, you'd have your desired equality. Feb 20, 2014 at 20:51

## 1 Answer

The answer to your question is YES, and this is easy to see if you observe that

$$f(x_{0}+) = f(x_{0}-) \neq f(x_{0})$$

implies the difference quotients based at $x = x_{0}$ become unbounded as you approach $x_{0}.$ That is,

$$\lim_{x \rightarrow x_{0}}\left|\frac{f(x) - f(x_{0})}{x - x_{0}}\right| = \infty,$$

which is due to the fact that for $x \rightarrow x_{0}$ we have

$$|x - x_{0}| \rightarrow 0 \;\;\;\;\; \text{and} \;\;\;\;\; |f(x) - f(x_{0})| \;\; \rightarrow \;\; |f(x_{0}+) - f(x_{0})| > 0$$