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Can anyone explain the uniform continuity clearly with picture if possible?? I have read the section on this topic in my text book but I am still not clear on this. Thanks.

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In my text book, it gives two definitions of a continuous function. One with sequence in the domain of a function and $\varepsilon$-$\delta$ definition. I think I am good with a continuous function at a point. The definition of the uniform continuity in my text book is this.

Let $f$ be a real-valued function defined on a set $S \subseteq R.$ Then $f$ is uniformly continuous on $S$ if $\forall \varepsilon >0,\exists \delta > 0\text{ s.t } x,y \in S \text{ and } |x-y| < \delta \Rightarrow \left|f(x)-f(y)\right|<\varepsilon.$

From this definiton the points $x,y$ are not fixed points in the domain. So for given $\varepsilon >0$ we find $\delta >0$ such that the distance between $x$ and $y$ in the domain is less than $\delta$ implies the distance between $f(x)$ and $f(y)$ less than given $\varepsilon.$ Then the function is uniformly continuous. This is as far as I know about the uniform continuity.

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  • $\begingroup$ are you familiar with definition of continuity? what did you understood... what is the extra condition in definition of uniform continuity? $\endgroup$
    – user87543
    Nov 1, 2013 at 2:27
  • $\begingroup$ edited the post!! $\endgroup$
    – eChung00
    Nov 1, 2013 at 2:44

3 Answers 3

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First, uniform continuity is about an interval, not just a point. You can say f(x) is continuous on the interval $(a,b)$ means: $\lim_{x \rightarrow c} f(x) = f(c)$ for all $c$ in $(a,b)$. Then you have to say what you mean by a limit, and you get the old $\varepsilon$-$\delta$ statement: for every $\varepsilon$ there exists a $\delta$ such that $|x-c| < \delta \Rightarrow |f(x) - f(c)| < \varepsilon$.

Now what they never tell you is that $\delta$ depends on $x$. For some functions $\delta$ has to get smaller and smaller to get $|f(x) -f(c)| < \varepsilon$. An example of this is the function $f(x) = 1/x$ on $(0,1)$. It's continuous there everywhere, but as $x$ approaches $0$ the function gets steeper and steeper. That means you have to take your $x$ and $c$ closer and closer together -- i.e $\delta$ smaller and smaller, to get $|f(x) -f(c)| < \varepsilon$.

What uniform continuity means is that the $\delta$ does NOT depend on x. You can find a single $\delta$ that depends only on $\epsilon$ for the entire interval.

Intuitively uniform continuity means your function can't get infinitely steeper on $(a,b)$ the way $1/x$ does on $(0,1)$. Steepness doesn't only mean that $f$ may go off to infinity in your interval. It could also oscillate around in some unpleasant way. Look for example at $f(x) = \sin(1/x)$ on $(0,1)$.

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  • $\begingroup$ Thanks for your comment. The example you gave is same example in my text book. There is another example with $f(x) = \frac{1}{x}$ on $[a,\infty)$ where $a>0.$ If I understood correctly, the function $f$ on $(0,1)$ is not unformly continous because the $\delta$ is depends on $x$, right?? But the second example is uniformly continuous based on the example in the book. why is it?? Since $a>0$ I think $(0,\infty)$ is same interval with $[a,\infty).$ $\endgroup$
    – eChung00
    Nov 1, 2013 at 2:51
  • $\begingroup$ Note, however, that you shouldn't take the "steepness" analogy to get the idea that the derivative has to be bounded. Take, for example, $f(x)=\sqrt x$ on $[0,1]$, which is uniformly continuous, but has unbounded derivative. $\endgroup$ Nov 1, 2013 at 3:01
  • $\begingroup$ @eChung00 For your example of $1/x$ on $[a,\infty)$, you don't need arbitrarily small $\delta$, because the function gets flatter as $x$ gets larger. Indeed, $\dfrac1x-\dfrac1y=\dfrac{y-x}{xy}\le\dfrac{y-x}{a^2}$ for all $x,y\in[a,\infty)$, so since I can bound the difference by a function of $y-x$ (that goes to $0$), independent of $x$ and $y$ individually, the function is uniformly continuous. The converse to my previous comment is true, though; a bounded derivative implies uniform continuity, which can be helpful, as in this case, where $|f'|\le a^{-2}$ on the whole range. $\endgroup$ Nov 1, 2013 at 3:09
  • $\begingroup$ What also implies uniform continuity is that the function is continuous on a compact set -- which in this case you can think of as one or several closed intervals. That's the way to prove what Mario is saying about $f(x) = \sqrt x$. $\endgroup$
    – Betty Mock
    Nov 1, 2013 at 20:09
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One intuitive way to think about it: if a function is uniformly continuous, as long as the "distance" between "$a$" and "$b$" is within some number $\delta$, you are guaranteed that the "distance" between $f(a)$ and $f(b)$ will be within some number $\varepsilon$. The rigorous correct way to think about it is the other way around. You first fix $\varepsilon$ and try to find whether you can find such $\delta$.

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  • $\begingroup$ Also, note that $a$ and $b$ need to be in the domain of $f$ $\endgroup$ Nov 1, 2013 at 6:53
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Uniform continuity gives you a continuous function for all $x$ in your domain, while pointwise continuity does not guarantee this. For example, prove that the sequence of functions $f_n(x) = x^n$ for $x \in [0,1]$ is pointwise continuous, but not uniformly continuous. This picture should give you an idea of the behavior for each $f_n$ enter image description here

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  • $\begingroup$ This answer is not correct or at least is confusing. I think your point is that the sequence $f_n$ is not uniformly convergent , only pointwise convegent. This is a differente concept. Each $f_n$ is uniformly continuous (since they are continuous functions on compact intervals) $\endgroup$ Apr 8, 2023 at 11:30

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