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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).

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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 $0$.

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  • $\begingroup$ "There is a closed interval $I$ containing $x_0$ on which $f$ is continuous", I do not agree. take for example : $ f(x<0)=0, f(x=0)=\frac{1}{2}, f(x>0)=1 $, there is no closed interval containing x=0 on which $f$ is continuous. $\endgroup$ – User Dec 17 '19 at 9:55
  • $\begingroup$ @User Given that in that case, the function that is only nonzero at one point becomes a counterexample to the problem adressed by the OP, I think it's safe to assume that the task does not consider functions such as yours to be piecewise continuous. $\endgroup$ – 5xum Dec 17 '19 at 13:12
  • $\begingroup$ Thanks for responding, of course it's definition dependent. Tut I'm pretty sure the Heaviside Step Function (same function from previous comment) is a piecewise continuous function using every definition I've ever seen. If I understand the question correctly, it's also not a counter example, please see my answer math.stackexchange.com/a/3481585/61487 . $\endgroup$ – User Dec 19 '19 at 8:12
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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 :)

enter image description here

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  • $\begingroup$ "I can select a continuous closed interval, which contains $x_0$", no you can't, see my comment on math.stackexchange.com/a/672048/61487 . In addition, the plotted function doesn't hold $f(x) = \frac{1}{2}[f(x-) +f(x+)]$ $\endgroup$ – User Dec 17 '19 at 9:58
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For every $x_0 \in (a,b)$ both limits exists (but do not necessarily equal):

$$\lim_{\Delta x \to 0} f(x_0-\Delta x)$$ $$\lim_{\Delta x \to 0} f(x_0+\Delta x)$$

At least one of the limits isn't equal to $0$, and that's the side of $x_0$ where there is an open interval I on which $f(x) \neq0$ for all $x \in I$.

Define $I_{x_0} = I \cup \{ x_0 \} $, an interval containing $x_0$. $f(x) \neq0$ for all $x \in I_{x_0}$.

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  • $\begingroup$ Given that you allow (in your comment to my answer) the function $f(x<0)=0, f(x=0)=\frac12, f(x>0)=1$, do you also allow the function $g(x<0)=0, g(x=0)=\frac12, g(x>0)=0$? If so, $g$ is a counterexample, since both limits are equal to $0$. $\endgroup$ – 5xum Dec 19 '19 at 8:18
  • $\begingroup$ Note: the point of my comment is that I believe OP is working with a definition of "piecewise continuous" that reads something like: A function is piecewise continuous on $(a,b)$, if there exist some numbers $x_i$, $a=x_0<x_1<\cdots<x_n=b$, such that for each $i$, $f$ is continuous on $(x_i, x_{i+1})$. In which case both of our answers are sufficient to answer the question. $\endgroup$ – 5xum Dec 19 '19 at 8:23

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