Question

Show that f:[0,1]_R defined by f(x)=x^2 is uniformly continuous on [0,1]

Answer 1

A function f is said to he uniformly continuous is for arbitrary points x and y,

|f(x)-f(y)|< $\in$ when |x-y|<$\delta$ such that $\delta$ is independent of x and y.

Here, f(x)= x² on [0,1]

Let y$\in$ [0,1] such that

for |x-y|<$\delta$

|f(x)-f(y)|= |x²-y²|

<|(x-y)(x+y)|

<2|x-y|

<2$\delta$

Choose $\delta=\in /2$

|f(x)-f(y)|<$\in$ for |x-y|<$\delta$ where $\delta$ is independent of x and y.

Hence, f is uniformly continuous on [0,1]