Why does piecewise continuity imply function is nonzero in some interval?

Question

My textbook says that

If $f(x)$ is piecewise continuous on $(a,b)$ and satisfies $f(x) = \frac{1}{2} [f(x_{-})+f(x_{+})]$ for all $x\in(a,b)$, and if $f(x_{0})\neq 0$, then $|f(x)|>0$ on some interval containing $x_{0}$.

Why is this true?

Edit:

My attempt to prove this claim:

Suppose $f \in PC (a,b), f (x) = \frac{1}{2} [f(x-) + f(x+)]$, and $f(x_{0}) \neq 0$ for some $x_{0} \in (a,b)$.

Case I: If $f$ is continuous at $x_{0}$, then by the definition of continuity, there is a neighborhood $U$ containing $x_{0}$ where $f(x)\neq0$ for $x \in U$.

Case II: If $x_{0}$ is a point of discontinuity, then because $f$ is piecewise continuous, the values $$f(x-)=\lim_{\varepsilon\to0+} f(x-\varepsilon), \quad f(x+) = \lim_{\varepsilon \to 0+} f(x+\varepsilon)$$

always exist. For fixed $x_{0} \in (a,b)$ such that $f(x_{0})\neq0$, the condition $f(x_{0}) = \frac{1}{2} [f(x_{0}^{-}) + f(x_{0}^{+})]$ implies that $f(x_{0}^{-})$ and $f(x_{0}^{+})$ are not simultaneously zero. Then:

II.i) if $f(x_{0}^{-})$ is zero, choose the interval $[x_{0},x_{0}+\varepsilon)$.

II.ii) if $f(x_{0}^{+})$ is zero, choose the interval $[x_{0}-\varepsilon, x_{0})$.

II.iii) if $f(x_{0}^{-}), f(x_{0}^{+})$ are not zero, choose the interval $(x_{0}-\varepsilon, x_{0}+\varepsilon)$.

Answer 1

When $\ f(x_0)>0\ $ then $\ f(x_-)> 0\ $ or $\ f(x_+) > 0. $ Thus, either $\ f\ $ is positive in an interval $\ (x_0-h;x_0)\ $ or in $\ (x_0;x_0+h)\ )$, respectively (or even for both), for certain $\ h>0.$

Remark 1: One may add $\ x_0\ $ to the considered intervals.

Remark 2: The result is more general, the assumption about piecewise continuity can be replaced by a weaker (and much more elegant). It is enough to assume that both limits $\ f(x_-)\ $ and $\ f(x_+)\ $ exist for every $\ x\in(a;b).$ And let's keep the other assumptions intact.

Answer 2

Observe that $\lvert f(x)\rvert>0$ and $f(x)\ne0$ are the same thing.

If $f$ is continuous at $x_0$, then by definition of continuity $\{x\in\Bbb R\,:\, f(x)\ne0\}$ must contain a neighbourhood of $x_0$.

If $f$ is not continuous at $x_0$, then there cannot be two sequences $(z_n)_{n\in\Bbb N}$ and $(y_n)_{n\in\Bbb N}$ such that $f(y_n)=f(z_n)=0$, $z_n\nearrow x_0$ and $y_n\searrow x_0$. In fact, if such were the case, then necessarily $f(x_0^+)=f(x_0^-)=0$, and therefore $f(x_0)=0$.

If there is no sequence $(y_n)_{n\in\Bbb N}$, then $\{x\,:\, f(x)\ne0\}$ contains an interval in the form $[x_0,x_0+\varepsilon)$ for some $\varepsilon>0$. If there is no sequence $(z_n)_{n\in\Bbb N}$, then it contains $(x_0-\varepsilon,x_0]$.