# Uniform continuity of function

Let $f:( 0, ∞)→R$ such that $f(x) = 1/x$ then we know it is not uniformly continuous. As shown in graph, for any given $ε >0$, the same $δ$ does not work everywhere on graph. That is, $δ$ changes for same $ε$ for the different portion of graph, So that function is not uniformly continuous.

Now, if we consider $g(x) = 1/x$ in $(a, ∞)$ where $a > 0$ then too, I have different $δ$, for same $ε$ in different portion of graph. So why $g(x) = 1/x$ is uniformly continuous on $(a,∞)$ where $a> 0$?

• Intuitively the problem with $f$ that makes it fail to be uniformly continuous that the slope of $f$ grows arbitrarily large. For any $a > 0$, the derivative of $g$ is bounded and so $g$ is uniformly continuous. See if you can work through the $\epsilon-\delta$ argument and you'll find there indeed is a $\delta$ for every $\epsilon$ which works for the entire interval. – eepperly16 May 16 '17 at 5:45
• Please clarify, how I get same $δ$ for $g(x)$ which works for entire interval. – Akash Patalwanshi May 16 '17 at 5:49
• The $\delta$ is not unique. Choosing a smaller $\delta$ works. Uniform continuity means that we can choose a $\delta$ that works regardless of which points are chosen - dependent only on $\epsilon$. – Manuel Guillen May 16 '17 at 5:58

If we consider the interval $I=(a,\infty)$, then we see that, for any $x,x' \in I$: \begin{align*} \lvert f(x) - f(x') \rvert &= \left\lvert \frac{1}{x} - \frac{1}{x'} \right\rvert \\ &= \left\lvert \frac{x'-x}{x\cdot x'} \right\rvert \\ &< \frac{1}{a^2} \cdot \left\lvert x'-x \right\rvert \\ \end{align*} For any $\epsilon > 0$, we can choose $\delta = a^2\epsilon$ and see that: $$\lvert x - x' \rvert < \delta \implies \lvert f(x) - f(x') \rvert < \epsilon \qquad \forall x,x' \in I$$ By definition, $f(x)$ is uniformly continuous on $I$.

To answer your question about why concerning the $\delta$ - the $\delta$ is not unique. Choosing a smaller $\delta$ works. Uniform continuity means that we can choose a $\delta$ that works regardless of which points are chosen - dependent only on $\epsilon$.

• Thanks. Can we see this $δ = a^2 ε$ works for entire interval via graph? – Akash Patalwanshi May 16 '17 at 6:10

Let $\epsilon > 0$ and $g : (a,\infty) \to \mathbb R$ given by $g(x) = 1/x$. Take $\delta = a^2\epsilon$. Note that for any $x,y \in (a,\infty)$, $xy \ge a^2$ so $1/(xy) \le 1/a^2$. Then if $|x-y| < \delta$,

$$\left| \frac{1}{x} - \frac{1}{y} \right| = \frac{|y-x|}{xy} \le \frac{|y-x|}{a^2} < \frac{\delta}{a^2} = \epsilon$$

• Thanks get it. I think it is impossible to see this δ works for entire interval via graph? – Akash Patalwanshi May 16 '17 at 6:06
• @AkashPatalwanshi It is: the problem that you had originally is the closer you get to 0, the larger your $\delta$ needs to be. In this case, you have a hard boundary on the left, at $x = a$: the graph is at its steepest there so if you have a $\delta$ that works in that region, it works everywhere. In other words: if you have an $(\delta, \epsilon)$ rectangle that works near $x = a$, you can use the same rectangle everywhere and since the graph is less steep there, it will definitely fit in the same rectangle. – CompuChip May 16 '17 at 8:34

Suppose it is not uniformly cont on $(a,\infty)$. Then there exist sequence $u_n$ and $v_n$ such that $|u_n-v_n|\to 0$ but $|f(u_n)-f(v_n)|\not\to 0$. However we have $|f(u_n)-f(v_n)| =|1/u_n -1/v_n| =|\dfrac{v_n-u_n}{u_nv_n}|$. As long as $u_nv_n$ doesn't go to zero(because of our interval), we can evaluate the limit of the top and bottom separately. The numerator limits to zero, and the denominator will be bounded by $1/a$ and $0$. Thus $|f(u_n)-f(v_n)|\to 0$. This is a contradiction, thus $f$ is uniformly continuous on the interval.

Intuitively, uniform continuity means that the function cannot get too steep. As slope increases, the ratio of change in y to change in x increase. In epsilon-delta terminology, this means that our function cant have an infinitely large epsilon for a infinitely small delta. In this case the "steep" part of the function happens in the limit towards zero. When we choose the interval to $(0,\infty)$ we lose uniform continuity. But when we cap the interval at $a$, the slope of our function can be bounded, and so it is uniformly continuous.