I have some questions on continuity. What is the difference between continuous and uniformly continuous function? Please explain with this question. Find $f(x)=x^2$ is uniformly continous on [0,$\infty$[ ?. Is there any particular method to follow?


The difference is at the "$\delta$" level in the $\epsilon$-$\delta$ definition of continuity.

Let $f$ be a function on $(a,b)\subseteq \mathbb R$. Then $f$ is continuous on $(a,b)$ if

$\forall x_0\in(a,b),~ \forall \epsilon>0~~\exists \delta=\delta(\epsilon, x_0)~:~\forall x~:~|x-x_0|\Rightarrow |f(x)-f(x_0)|<\epsilon,$

while $f$ is uniformly continuous on $(a,b)$ if

$\forall x_0\in(a,b),~ \forall \epsilon>0~~\exists \delta=\delta(\epsilon)~:~\forall x~:~|x-x_0|\Rightarrow |f(x)-f(x_0)|<\epsilon.$

In the case of uniform continuity the parameter $\delta$ is independent of the chosen point $x_0$ in the domain $(a,b)$. Note that uniform continuity makes sense only on (sub)sets, while continuity can be defined on each point of a given (sub)set, as usual.

Moreover, each uniformly continuous function $f$ on a subset $(a,b)\subset \mathbb R$ is ivi continuous. The converse does not hold. An example:

  • $f(x)=x^2$ is uniformly continuous on $(0,1)$

Let $\epsilon >0$, $x,x_0\in (0,1)$ and $|x-x_0|<\delta,$ for some $\delta$ to be found. We want to show that $f$ is continuous and $\delta$ can be chosen s.t. $\delta=\delta(\epsilon)$ (no dependence w.r.t. $x_0$).


$$|f(x)-f(x_0)|=|x-x_0||x+x_0|<\delta |x+x_0|;$$

using the triangular inequality $|x+x_0|\leq |x|+|x_0|$ and $x,x_0\in (0,1)$ we can write

$$|f(x)-f(x_0)|=|x-x_0||x+x_0|<\delta |x+x_0|\leq\delta(|x|+|x_0|)<2\delta;$$

then, choosing $\delta :=\frac{\epsilon}{2}$ we show uniform continuity of $f$ on $(0,1)$.

  • $\begingroup$ Please, let me know if you need more details concerning $f(x)=x^2$ $\endgroup$ – Avitus Jul 26 '14 at 9:31

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