$\lim_{x \to 0+} w(a, \delta) = 0$ iff f is continuous at $a$

Question

Let $f$ be a bounded function on $\Bbb R$ and $a\in \Bbb R$. For $\delta > 0$, let $w(a, \delta) = Sup|f(x) - f(a)|$, $x \in [a - \delta , a + \delta]$. Then prove:

1. $w(a, \delta_1) \leq w(a, \delta_2)$ if $\delta_1 \leq \delta_2$.

2. $\lim_{\delta \to 0+} w(a, \delta) = 0$, $\forall a \in R$.

3. $\lim_{\delta \to 0+} w(a, \delta)$ need not exist.

4. $\lim_{\delta \to 0+} w(a, \delta) = 0$ iff f is continuous at $a$

My Attempt: Let $f$ is continuous at $a$ iff $\lim_{x \to a} f(x) = f(a)$ iff $\lim_{x \to a} (f(x) - f(a)) = 0$ iff supremum of $|f(x) - f(a)|$ tends to $0$ as x tends to $a$. I have no idea how to Prove or Disprove other options. Please help me.

Answer 1

Proof of 4): Suppose $\omega(a,\delta) \to 0$ as $\delta \to 0+$. Let $\epsilon >0$. There exists $r$ such that $0<\delta<t$ implies $\omega(a,\delta)<\epsilon$. This implies $|f(x)-f(a)| <\epsilon$ whenever $|x-a| <\delta$. Hence $f$ is continuous at $a$.

Suppose $f$ is continuous at $a$. Let $\epsilon >0$. There exists $r>0$ such that $|f(x)-f(a)| <\epsilon$ whenever $|x-a| \leq r$. This implies that $\omega(a,r) \leq \epsilon$. In fact $\omega(a,\delta) \leq \epsilon$ whenever $\delta <r$. Hence $\omega(a,\delta) \to 0$ as $\delta \to 0+$.