We consider real-valued functions.

A real-valued function \(f : I \to \mathbb R\) (where \(I \subseteq\) is an interval) is continuous at \(x_0 \in I\) when: \[(\forall \epsilon > 0) (\exists \delta > 0)(\forall x \in I)(\vert x- x_0 \vert \le \delta \Rightarrow \vert f(x)- f(x_0) \vert \le \epsilon).\] When \(f\) is continuous at all \(x \in I\), we say that \(f\) is continuous on \(I\).

\(f : I \to \mathbb R\) is said to be uniform continuity on \(I\) if \[(\forall \epsilon > 0) (\exists \delta > 0)(\forall x,y \in I)(\vert x- y \vert \le \delta \Rightarrow \vert f(x)- f(y) \vert \le \epsilon).\]

Obviously, a function which is uniform continuous on \(I\) is continuous on \(I\). Is the converse true? The answer is negative.

### An (unbounded) continuous function which is not uniform continuous

The map \[

\begin{array}{l|rcl}

f : & \mathbb R & \longrightarrow & \mathbb R \\

& x & \longmapsto & x^2 \end{array}\] is continuous. Let’s prove that it is not uniform continuous. For \(0 < x < y\) we have
\[\vert f(x)-f(y) \vert = y^2-x^2 = (y-x)(y+x) \ge 2x (y-x)\] Hence for \(y-x= \delta >0\) and \(x = \frac{1}{\delta}\) we get

\[\vert f(x) -f(y) \vert \ge 2x (y-x) =2 > 1\] which means that the definition of uniform continuity is not fulfilled for \(\epsilon = 1\).

For this example, the function is unbounded as \(\lim\limits_{x \to \infty} x^2 = \infty\). Continue reading Continuity versus uniform continuity