Answer to Question #95328 in Real Analysis for Rajni

Question

Show that the function f:]-1,1] [ --> R given by f(x) =x^3 is uniformly continuous and deduce that f is continuous at the point zero?

Answer 1

Remember that

y= f(x) is uniformly continuous  on ]-1,1] if $\forall \epsilon >0 \forall x,y \in ]-1,1] \vert x-y\vert< \delta \Rightarrow \vert f(x)-f(y)\vert< \epsilon$ .

If $\forall x,y \in ]-1,1]$ :

$|x^3 - y^3| = |(x - y) ( x^2 + yx + y^2)|= \\=|x - y| | x^2 + yx + y^2| \leq\\\leq |x - y|( x^2 + | yx| + y^2) <\delta |1+1+1| =3\delta$

Proof.

Let $\epsilon >0 .$ Choose $\delta= \frac{\epsilon}{3}$  .

Then $\forall x,y \in ]-1,1]$ with $\vert x-y\vert< \delta$ we have

$|x^3 - y^3| = |(x - y) ( x^2 + yx + y^2)|= \\=|x - y| | x^2 + yx + y^2| \leq \\ \leq |x - y|( x^2 + | yx| + y^2) <\delta |1+1+1| =3\delta=3\frac{\epsilon}{3}=\epsilon\\$ .

If $\vert x-y\vert< \delta$ :$|x^3 - y^3| <\epsilon$ .

So

$f(x)=x^3$ is uniformy continuous on ]-1,1] .

Since f(x) is uniformy continuous ]-1,1]  then it  is continuous at the point zero (x=0).