# a very basic question on uniform continuity of function

A very basic and very strange question came to me.

Let $D\subseteq\mathbb {R}$, $f:D\to\mathbb{R}$ a continuous function. Then $f$ is uniformly continuous on $D$ iff. $$\forall\epsilon>0\ \exists\delta>0\,\,\text{such that}\,\,\forall\,x,y\in D$$ we have $$|y-x|<\delta\,\Longrightarrow\, |f(y)-f(x)|<\epsilon$$. Now, let $f\in\mathbb{R}$ be uniformly continuous on $D$. Show that for every $\epsilon>0$, there exists $\delta>0$ such that $$|f(y)-f(x)|<\frac{\epsilon}{2},\quad\text{whenever}\quad |x-y|<\frac{\delta}{2}$$.

I really do not know what need to be proven but I was asked to write a proof on this.

Thanks for any helps.

• How do you define uniform continuity here? – Weaam Oct 4 '15 at 12:02
• Where did you get stuck? If $\epsilon>0$ then also $\frac \epsilon2>0$. If $\delta_0>0$ then also $\delta=2\delta_0>0$ – Hagen von Eitzen Oct 4 '15 at 12:02
• What does it mean $f\in\mathbb R$ ? – Svetoslav Oct 4 '15 at 12:02
• I admire the person who posed the question. You might try proving the following first: Suppose $f$ is continuous at $a$. show that for any $\epsilon > 0$, there's a $\delta > 0$ such that $|x - a| < \delta$ implies $|f(x) - f(a)| < \epsilon/2$. The you might try showing that there's $\delta$ where $|x - a| < \delta$ implies $f(x) - f(a) < \epsilon/2$. (Note absence of absolute values here). By that time, you'll be getting the general idea. – John Hughes Oct 4 '15 at 12:02
• @JohnHughes good one :) – Svetoslav Oct 4 '15 at 12:03

Replace $\epsilon ,\delta$ in the definition of uniform continuity by $\epsilon^* ,\delta^* .$ Write it out. Now,given $\epsilon >0 ,$ let $\epsilon^*=\frac {1}{2}\epsilon$ and take $\delta^* >0$ that satisfies the statement in the definition. Now let $\delta^* =\frac{1}{2}\delta.$ As an exercise, describe what uniform continuity means without mentioning our friends epsilon and delta.
• Dear user254665, is this what you mean? If not, please correct me. (Anyone, please) $f$ is uniform continuous, then given any $\epsilon^*>0$, there exists a $\delta^*>0$ such that for any $y$, $x\in D$, we have $$|y-x|<\delta^*\,\,\Longrightarrow\,\,|f(y)-f(x)|<\epsilon^*$$. Now take $\epsilon^*=\frac{1}{2}\epsilon$. Let $\delta^*=\frac{\delta}{2}$ Then for any $$x,y\in D,\,\,|y-x|<\delta^* \,\,\Longrightarrow |y-x|<\frac{\delta}{2}$$, which implies $|f(y)-f(x)|<\frac{\epsilon}{2}$. – math101 Oct 4 '15 at 13:53
• Exactly right. Consider the definition of continuity at one point $x$.The idea of uniform continuity is that given $\epsilon^*$, there is a choice of $\delta^*$ that works for all $x$. – DanielWainfleet Oct 4 '15 at 18:06
Since $f$ is uniform continuous, then given any $\epsilon^*>0$, there exists a $\delta^*>0$ such that for any $y$, $x\in D$, we have $$|y-x|<\delta^*\,\,\Longrightarrow\,\,|f(y)-f(x)|<\epsilon^*$$. Now take $\epsilon^*=\frac{1}{2}\epsilon$. Let $\delta^*=\frac{\delta}{2}$ Then for any $$x,y\in D,\,\,|y-x|<\delta^* \,\,\Longrightarrow |y-x|<\frac{\delta}{2}$$, which implies $|f(y)-f(x)|<\frac{\epsilon}{2}$.
• That looks like a correct proof. What my hints were meant to get at was the idea that the "epsilon" in the definition didn't have to be the one that you were given in some situation where you needed to apply continuity. To do the 2nd problem in my hint, you need only observe that $f(x) - f(a) \le |f(x) - f(a)|$, because every number is no greater than its absolute value. My idea was to make you really think about every aspect of the definition of continuity, since there wasn't much else to your question. – John Hughes Oct 4 '15 at 22:22