I was working on exercise 4.1 of Rudin's Principles of Mathematical Analysis by Rudin which states
Suppose $f$ is a real function defined on $\mathbb{R}^1$ which satisfies
$$\lim_{h\rightarrow 0}[f(x+h)-f(x-h)]=0$$
for every $x\in\mathbb{R}^1$. Does this imply that $f$ is continuous?
I now know that the answer is in fact "no" and understand the given proof for why. However I'm having a lot of trouble figuring out where I went wrong in my proof that the statement does imply that $f$ is continuous:
Let $$g(h)=f(p+h)-f(p-h)$$ By the definition of a limit we know that
for each $p\in \mathbb{R}$ for all $\epsilon_0>0$ there exists a
$\delta_0>0$ such that $$d_Y(g(h),0)<\epsilon_0$$ for all $h$ where
$$0<d_X(h,0)<\delta_0$$ We wish to show that for each $p\in\mathbb{R}$
for all $\epsilon>0$ there exists a $\delta>0$ such that
$$d_Y(f(x),f(p))<\epsilon$$ for all $x\in \mathbb{R}$ for which
$$0<d_X(x,p)<\delta$$ Let us consider \begin{align}
d_Y(g(h),0)&=|f(p+h)-f(p-h)|\\
&=|f(p+h)+f(p)-f(p)-f(p-h)|\\
&\leq |f(p+h)-f(p)|+|f(p-h)-f(p)|\\
&=d_Y(f(p+h),f(p))+d_Y(f(p-h),f(p))\\
&\textrm{If }d_Y(f(p+h),f(p))\geq d_Y(f(p-h),f(p))\textrm{ then}\\
&\leq2d_Y(f(p+h),f(p))\leq \epsilon_0\\
&\textrm{Else}\\
&\leq2d_Y(f(p-h),f(p))\leq\epsilon_0\\ \end{align} We find this relation to be true for all $h$ such that $$0<|h|<\delta_0$$ We can
see that this corresponds to the following statement:For each $p\in\mathbb{R}$ for all $\epsilon_0>0$
$$2d_Y(f(x),f(p))\leq \epsilon_0 $$ is true for all $x\in\mathbb{R}$ for
which $$0<d_X(x,p)<\delta_0$$ This is the statement that $f$ is
continuous at all point $p\in\mathbb{R}$.
I was wondering if someone could help we locate where I went awry
Best Answer
Can you find a counterexample to the question? What function satisfies the conditions but isn't continuous?
If you consider the counterexample while reading your proof, you will find that this step is suspect at the point of discontinuity ($ p = 0$ in my example).
(Lots of abuse of $\epsilon$ notation used here. You may treat them as mostly distinct.)
You have no information about $|f(p+h)-f(p)|$, and couldn't draw that conclusion (in the last 2 lines).
You are wanting to show (in the previous line of your proof), that this value is $ \leq \epsilon$.
In fact, referring to my example, we had $|f(0+h) - f(0)| = 1$, and cannot bound it by $\epsilon$.
You ended up assuming that it was $ \leq \epsilon$ (in the last 2 lines) in order to prove that it was $ \leq \epsilon$.