Let $f: R \to R$ be a function. Define $x \in R$ and $n \in Z^+$ and
$\phi_n(x) = \sup \big\{ |f(s) - f(t)| : s,t \in (x-\frac{1}{n}, x + \frac{1}{n}) \big\} $
Prove that the sequence ${\phi_n}$ converges uniformly to $0$ if and only if $f$ is uniformly continuous.
Here is what I did:
Proof: ($\Rightarrow$) Suppose the sequence $\{ \phi_n \}$ converges uniformly to $0$. Then for $\epsilon > 0$, $n > N(\epsilon)$ imply
$| \phi_n(x)- 0| = \sup \big\{ |f(s) - f(t)| : s,t \in (x-\frac{1}{n}, x + \frac{1}{n}) \big\} < \epsilon $
So for all $ s,t \in (x-\frac{1}{n}, x + \frac{1}{n})$ , $ | s- t| < \delta < \frac{2}{n} $ then $|f(s) - f(t)| < \epsilon$. Hence, $f$ is uniformly continuous.
($\Leftarrow$)Now suppose that $f$ is uniformly continuous. Then for each $\epsilon > 0$, there exist a $\delta > 0$ such that for all $s ,t \in (x-\frac{1}{n}, x + \frac{1}{n}) $ we have $|f(s) - f(t)| < \epsilon$. Then
$| \phi_n(x)| = \sup \big\{ |f(s) - f(t)| : s,t \in (x-\frac{1}{n}, x + \frac{1}{n}) \big\} < \epsilon $.
Thus $\phi_n$ converges to $0$ uniformly.
Can you point out if there is a mistake and help me to fix it if that's possible. Thank you!
The proof of the forward implication requires some modification.
Since $\phi_n \to 0$ uniformly, for any $\epsilon >0$ there exists $N_\epsilon$ such that if $n \geqslant N_\epsilon$ then for every $x \in \mathbb{R}$ we have $\phi_n(x) < \epsilon$. This implies that for any given $x$, if $s,t \in (x - 1/N_\epsilon, x + 1/N_\epsilon)$ then $|f(s) - f(t)| \leqslant \phi_{N_\epsilon}(x) < \epsilon$.
Consider any $s,t \in \mathbb{R}$. Take $x = s$ and $\delta_\epsilon = 1/N_\epsilon$. Then if $|t-s| < \delta_\epsilon$, we have both $s,t \in (s - 1/N_\epsilon, s + 1/N_\epsilon)$ and $|f(s) - f(t)| < \epsilon$.
Based on what was used for this argument, you should be able to fix your proof of the reverse implication.