How to prove the following problem:
Suppose $f \in PC(a,b)$, where $PC(a,b)$ means the set of piecewise continuous functions on the interval $[a,b]$ and $f(x) = \frac{1}{2}[f(x-) +f(x+)]$ for all $x \in (a,b)$. Show that if $f(x_0) \neq 0$ at some point $x_0 \in (a,b)$, then $f(x) \neq 0$ for all $x$ in some interval containing $x_0$. ($x_0$ may be an endpoint on the interval).
Hint: There is a closed interval $I$ containing $x_0$ on which $f$ is continous. Use continuity of $f$ in $I$ to prove that values near $x_0$ produce function values near $f(x_0)$, therefore away from $0$.
Is this sufficient for an answer (using @5xum's hint):
I can select a continuous closed interval, which contains $x_0$, say $[i , j ]$. Because $f$ is continuous on that interval I know that:
$$\lim_{\Delta x \to 0} f(x_0-\Delta x) = \lim_{\Delta x \to 0} f(x_0+\Delta x) = f(x_0) \neq0$$
In addition, I can select interval $[i,j]$ to be arbitrarily small, so that for any $x \in [i,j]$ ($x \neq x_0$) it holds:
$$0 <|f(x )-f(x_0)| <\epsilon <|f(x_0)|$$
for some $\epsilon > 0$. This means that I can select such an interval (where $x_0\in[i,j]$) for which it holds that $f(x) \neq0$ for all $x \in [i,j]$. In the border cases I need to use only one of the left- or right-hand limits.
Please let me know if this is flawed :)
