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# Analysis Proof

We will prove:

The expression: $$\displaystyle \lim_{h\to 0}\left[f(x+h) - f(x-h) \right] = 0$$ $$\forall$$ $$x$$ $$\in$$ $$\mathbb{R}$$

does not imply continuity for real-valued functions.

Proof:

Consider the following function:

$$f(x) = \alpha$$ if $$x \in \mathbb{Z}$$

$$f(x) = 0$$ if $$x \notin \mathbb{Z}$$

where $$\alpha \in \mathbb{R}$$, $$\alpha \neq 0$$. Now let us examine the expression in question for $$f(x)$$:

If $$x \in \mathbb{Z}$$: $$\displaystyle \lim_{h\to 0}\left[f(x+h) - f(x-h) \right] = \alpha - \alpha =0$$

If $$x \notin \mathbb{Z}$$: $$\displaystyle \lim_{h\to 0}\left[f(x+h) - f(x-h) \right] =0-0=0$$

Clearly, $$f(x)$$ satisfies the expression $$\forall$$ $$x \in \mathbb{R}$$, since $$\mathbb{Z}\cup \mathbb{Z}^c = \mathbb{R}$$, where $$\mathbb{Z}^c$$ denotes the topological compliment of the set of all integers. It is now sufficient to show that $$f(x)$$ is not continuous on $$\mathbb{R}$$.

Since $$\mathbb{R}$$ is a closed set, we may bypass the metric definition of continuity and utilize an alternative definition:

"For any $$p \in \mathbb{R}$$, a function $$f(x)$$ is continuous at $$p$$ if and only if $$\displaystyle \lim_{x\to p} f(x) = f(p)$$."

Now, consider any interval $$I= (a,b) \subset \mathbb{R}$$ such that:

(1) $$a \notin \mathbb{Z}$$

(2) $$b \notin \mathbb{Z}$$

(3) $$\exists$$ $$c$$ such that $$c \in \mathbb{Z}$$ and $$a<c<b$$

(4) Only one such $$c$$ exists in $$I$$

$$\forall$$ $$x$$ such that $$a<x<c$$, any sequence {$$t_n$$} in $$(a,c)$$ such that $$t_n \rightarrow x$$ will have $$\displaystyle \lim_{n\to \infty}f(t_n) =0$$ because $$\forall$$ $$n$$, $$t_n \notin \mathbb{Z}$$.

Hence $$f(x-) = 0$$ $$\forall$$ $$a<x<c$$. The same can be said for any sequence in $$(c,b)$$, hence $$f(x+)=f(x-) =0$$ $$\forall$$ $$x \in I$$. However, let $$x=c$$. The we have:

$$f(c+)=f(c-) =0$$ and $$f(c) =\alpha$$, because $$c \in \mathbb{Z}$$.

This is a discontinuity, and since $$I$$ was arbitrary, we can generalize $$I$$ to the entire real line if we remove restriction (4).

Then $$f(x)$$ is not continuous at any $$x \in \mathbb{Z}$$ and hence not continuous on the real line, even though $$f(x)$$ satisfies:

$$\displaystyle \lim_{h\to 0}\left[f(x+h) - f(x-h) \right] = 0$$ $$\forall$$ $$x$$ $$\in$$ $$\mathbb{R}$$

This is a counterexample to the assumption that the above statement implies continuity, hence we have proven that said statement does not imply continuity.

QED

When I originally discovered this expression, I sought to prove that it did imply continuity. After many failed attempts, I searched for a counterexample, arriving at the above.

Note by Ethan Robinett
2 years, 2 months ago