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.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

There are no comments in this discussion.