Uniform Continuity

Uniform continuity demands one $\delta$ that works at every point $x$ — $\delta$ may depend on $\varepsilon$ alone, never on $x$. Watch the required $\delta$ shrink as $x$ grows for $f(x) = x^2$, so no single $\delta$ covers all of $\mathbb{R}$.

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• Completeness and Compactness

Examples & Exercises

Prove a Lipschitz function (some $L$ with $\lvert f(x)-f(y)\rvert \leq L\lvert x-y\rvert$) is uniformly continuous.

$$\delta = \varepsilon / L$$

Show proof

1. Let $f$ be Lipschitz with constant $L \geq 0$. If $L = 0$, then $f$ is constant and any $\delta > 0$ works. Assume $L > 0$.
2. Given $\varepsilon > 0$, put $\delta = \varepsilon/L$ — $\delta$ depends on $\varepsilon$ alone, never on $x$.
3. For any $x, y$ with $\lvert x - y\rvert < \delta$ we have $\lvert f(x) - f(y)\rvert \leq L\lvert x - y\rvert < L\delta = \varepsilon$.
4. Hence the same $\delta$ works at every point $x$, so $f$ is uniformly continuous.

Prove $f(x)=x^2$ is uniformly continuous on $[0,1]$.

$$\lvert x^2-y^2\rvert = \lvert x-y\rvert\lvert x+y\rvert \leq 2\lvert x-y\rvert$$

Show proof

1. Let $x, y \in [0,1]$. Then $\lvert x^2 - y^2\rvert = \lvert x - y\rvert \lvert x + y\rvert$, and since $x + y \leq 2$, we get $\lvert x^2 - y^2\rvert \leq 2\lvert x - y\rvert$ — i.e. $x^2$ is Lipschitz on $[0,1]$ with constant $2$.
2. Given $\varepsilon > 0$, put $\delta = \varepsilon/2$ — independent of $x$. If $\lvert x - y\rvert < \delta$, then $\lvert x^2 - y^2\rvert \leq 2\lvert x - y\rvert < 2\delta = \varepsilon$.
3. Hence $f(x) = x^2$ is uniformly continuous on $[0,1]$. — on a bounded interval the "slope factor" $\lvert x + y\rvert$ is capped, so one $\delta$ suffices.

Prove $f(x)=x^2$ is not uniformly continuous on $\mathbb{R}$.

$$x_n = n,\quad y_n = n + \delta/2$$

Show proof

1. Take $\varepsilon_0 = 1$. Let $\delta > 0$ be arbitrary — a $\delta$ depending only on $\varepsilon_0$ would have to work for every pair of points.
2. Choose an integer $n > 1/\delta$ and put $x = n$, $y = n + \delta/2$. Then $\lvert x - y\rvert = \delta/2 < \delta$.
3. But $\lvert x^2 - y^2\rvert = \lvert y - x\rvert \lvert y + x\rvert = \frac{\delta}{2}\left(2n + \frac{\delta}{2}\right) = n\delta + \frac{\delta^2}{4} > 1 = \varepsilon_0$, since $n > 1/\delta$ gives $n\delta > 1$.
4. Hence no $\delta$ works for all pairs, so $f(x) = x^2$ is not uniformly continuous on $\mathbb{R}$. — the required $\delta$ tends to $0$ as $x$ grows.

3. Continuity