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I am trying to solve the following exercise in Rudin's "Rudin's Principles of Mathematical Analysis" book: (Ex 4.1)

Suppose $f$ is a real function defined on $R^1$ which satisfies $$\lim_{h\to 0}[f(x+h)-f(x-h)]=0$$ for every $x\in R^1$. Does this imply that $f$ is continuous?

The answer to this question is simply no, and it can be proved by using the function $f(x) = 1$ if $x\in \mathbb{Z}$ and $f(x) =0$, otherwise.

However, I am a little bit confused by this result, since I obtain the contrary by using the definitions of limits and continuity. In particular, I have the following derivation:

Define the function $$g(h) := f(x+h) - f(x-h)$$ for a fixed x. Then the hypothesis implies that $$\lim_{h\to 0}g(h)=0.$$ By using the definition of the limit, we have that $\forall \varepsilon >0$, $\exists \delta>0$, such that $|g(h)| < \varepsilon$ and $|h|< \delta/2$.

Now let $p, q \in \mathbb{R}$. Assume $p < q$, without loss of generality. Define $h := \frac{q-p}{2}$, $x := \frac{q+p}{2}$. Fix $\varepsilon >0$. Then, for $$|f(q)- f(p)| = |f(x+h)-f(x-h)| < \varepsilon,$$ we know that $$|q-p| = |x+h-x+h| = 2 |h| < \delta.$$ Hence, $f$ is continuous.

My question is: in which part of this derivation I am making a mistake?

Please note that there have been other questions regarding this exercise, for instance see this. However, I have a different concern, whose solution I think could be useful for other people.

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thmusic
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2 Answers2

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The error lies in “Hence, $f$ is continuous.” There is nothing before that assertion that justifies it. Proving that $f$ is continuous means proving that, for every $x\in\mathbb R$ and every $\varepsilon>0$, there is a $\delta>0$ such that $\lvert h\rvert<\delta\implies\bigl\lvert f(x+h)-f(x)\bigr\rvert<\varepsilon$, and you did not prove that.

José Carlos Santos
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    Def 4.5 in Rudin reads: $f$ is continuous at $p$ means $\forall \varepsilon$, $\exists \delta$, such that $|f(x) - f(p)|<\varepsilon$ for which $|x-p|<\delta$. I think the derivation I wrote agrees with this definition? – thmusic Mar 30 '19 at 21:35
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    It is the other way around: if $\lvert x-p\rvert<\delta$, then $\bigl\lvert f(x)-f(p)\bigr\rvert<\varepsilon$. – José Carlos Santos Mar 30 '19 at 21:42
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I think what is confusing you is that you showed something that looks like continuity, but is in fact not. To show that $f$ is continuous, we would have to show that $f$ is continuous at $x$ for each $x$. That would involve showing that $\pmb{|f(x+h)-f(x)|}$ can be made arbitrarily small by taking $h$ sufficiently small.

This is what your proof does not show: instead you demonstrated that $\pmb{|f(x+h)-f(x-h)|}$ can be made arbitrarily small by taking $h$ sufficiently small, which is actually just the condition $$\lim_{h\to0}[f(x+h)-f(x-h)] = 0.$$

Alex Ortiz
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