Show That 1 x is Uniformly Continuous
3.5: Uniform Continuity
- Page ID
- 49110
We discuss here a stronger notion of continuity.
Let \(D\) be a nonempty subset of \(\mathbb{R}\). A function \(f: D \rightarrow \mathbb{R}\) is called uniformly continuous on \(D\) if for any \(\varepsilon > 0\), there exists \(\delta > 0\) such that if \(u,v \in D\) and \(|u-v|<\delta\), then
\[|f(u)-f(v)|<\varepsilon .\]
Any constant function \(f: D \rightarrow \mathbb{R}\), is uniformly continuous on its domain.
Solution
Indeed, given \(\varepsilon > 0\), \(|f(u)-f(v)|=0<\varepsilon\) for all \(u,v \in D\) regardless of the choice of \(\delta\).
The following result is straightforward from the definition.
If \(f: D \rightarrow \mathbb{R}\) is uniformly continuous on \(D\), then \(f\) is continuous at every point \(x_{0} \in D\).
Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be given by \(f(x)=7 x-2\). We will show that \(f\) is uniformly continuous on \(\mathbb{R}\).
Solution
Let \(\varepsilon > 0\) and choose \(\delta=\varepsilon / 7\). Then, if \(u,v \in \mathbb{R}\) and \(|u-v|<\delta\), we have
\[|f(u)-f(v)|=|7 u-2-(7 v-2)|=|7(u-v)|=7|u-v|<7 \delta=\varepsilon \nonumber\]
Let \(f:[-3,2] \rightarrow \mathbb{R}\) be given by \(f(x)=x^{2}\). This function is uniformly continuous on \([-3,2]\).
Solution
Let \(\varepsilon > 0\). First observe that for \(u,v \in [-3,2]\) we have \(|u+v| \leq|u|+|v| \leq 6\). Now set \(\delta=\varepsilon / 6\). Then, for \(u,v \in [-3,2]\) satisfying \(|u-v|<\delta\), we have
\[|f(u)-f(v)|=\left|u^{2}-v^{2}\right|=|u-v||u+v| \leq 6|u-v|<6 \delta=\varepsilon. \nonumber\]
Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be given by \(f(x)=\frac{x^{2}}{x^{2}+1}\). We will show that \(f\) is uniformly continuous on \(\mathbb{R}\).
Solution
Let \(\varepsilon > 0\). We observe first that
\[\begin{align*} \mid \frac{u^{2}}{u^{2}+1} -\frac{v^{2}}{v^{2}+1} \mid &= | \frac{u^{2}\left(v^{2}+1\right)-v^{2}\left(u^{2}+1\right)}{\left(u^{2}+1\right)\left(v^{2}+1\right)} | \\[4pt] &= \frac{|u-v||u+v|}{\left(u^{2}+1\right)\left(v^{2}+1\right)} \leq \frac{|u-v|(|u|+|v|)}{\left(u^{2}+1\right)\left(v^{2}+1\right)} \\[4pt] &\leq \frac{|u-v|\left(\left(u^{2}+1\right)+\left(v^{2}+1\right)\right)}{\left(u^{2}+1\right)\left(v^{2}+1\right)} \\[4pt] &\leq|u-v|\left(\frac{1}{v^{2}+1}+\frac{1}{u^{2}+1}\right) \leq 2|u-v| , \end{align*}\]
(where we used that \(|x| \leq x^{2}+1\) for all \(x \in \mathbb{R}\), which can be easily seen by considering separately the cases \(|x|<1\) and \(|x| \geq 1)\).
Now set \(\delta=\varepsilon / 2\). In view of the previous calculation, given \(u,v \in \mathbb{R}\) satisfying \(|u-v|<\delta\) we have
\[|f(u)-f(v)|=\left|\frac{u^{2}}{u^{2}+1}-\frac{v^{2}}{v^{2}+1}\right| \leq 2|u-v|<2 \delta=\varepsilon . \nonumber\]
Let \(D\) be a nonempty subset of \(\mathbb{R}\). A function \(f: D \rightarrow \mathbb{R}\) is said to be Hölder continuous if there are constants \(\ell \geq 0\) and \(\alpha > 0\) such that
\[|f(u)-f(v)| \leq \ell|u-v|^{\alpha} \text { for every } u, v \in D .\]
The number \(\alpha\) is called Hölder exponent of the function. If \(\alpha = 1\), then the function \(f\) is called Lipschitz continuous.
If a function \(f: D \rightarrow \mathbb{R}\) is Hölder continuous, then it is uniformly continuous.
- Proof
-
Since \(f\) is Hölder conitnuous, there are constants \(\ell \geq 0\) and \(\alpha > 0\) such that
\[|f(u)-f(v)| \leq \ell|u-v|^{\alpha} \text { for every } u, v \in D . \nonumber\]
If \(\ell = 0\), then \(f\) is constant and, thus, uniformly continuous. Suppose next that \(\ell > 0\). For any \(\varepsilon > 0\), let \(\delta=\left(\frac{\varepsilon}{\ell}\right)^{1 / \alpha}\). Then, whenever \(u,v \in D\), with \(|u-v|<\delta\) we have
\[|f(u)-f(v)| \leq \ell|u-v|^{\alpha}<\ell \delta^{\alpha}=\varepsilon . \nonumber\]
The proof is now complete. \(\square\)
- Let \(D=[a, \infty)\), where \(a > 0\).(2) Let \(D=[0, \infty)\).
Solution
- Then the function \(f(x)=\sqrt{x}\) is Lipschitz continuous on \(D\) and, hence, uniformly continuous on this set. Indeed, for any \(u,v \in D\), we have
\[|f(u)-f(v)|=|\sqrt{u}-\sqrt{v}|=\frac{|u-v|}{\sqrt{u}+\sqrt{v}} \leq \frac{1}{2 \sqrt{a}}|u-v| ,\]
which shows \(f\) is Lipschitz with \(\ell=1 /(2 \sqrt{a})\).
Figure \(3.4\): The square root function.
- Then \(f\) is not Lipschitz continuous on \(D\), but it is Hölder continuous on \(D\) and, hence, \(f\) is also uniformly continuous on this set.
Indeed, suppose by contradiction that \(f\) is Lipschitz continuous on \(D\). Then there exists a constant \(\ell>0\) such htat
\[|f(u)-f(v)|=|\sqrt{u}-\sqrt{v}| \leq \ell|u-v| \text { for every } u, v \in D .\]
Thus, for every \(n \in \mathbb{N}\), we have
\[\left|\frac{1}{\sqrt{n}}-0\right| \leq \ell\left|\frac{1}{n}-0\right| .\]
This implies
\[\sqrt{n} \leq \ell \text { or } n \leq \ell^{2} \text { for every } n \in \mathbb{N} .\]
This is a contradiction. Therefore, \(f\) is not Lipschitz continuous on \(D\).
Let us show that \(f\) is Hölder continuous on \(D\). We are going to prove that
\[|f(u)-f(v)| \leq|u-v|^{1 / 2} \text { for every } u, v \in D .\]
The inequality in (3.9) holds obviously for \(u = v = 0\). For \(u > 0\) or \(v > 0\), we have
\[\begin{aligned}
|f(u)-f(v)| &=|\sqrt{u}-\sqrt{v}| \\
&=\left|\frac{u-v}{\sqrt{u}+\sqrt{v}}\right| \\
&=\sqrt{|u-v|} \frac{\sqrt{|u-v|}}{\sqrt{u}+\sqrt{v}} \\
& \leq \frac{\sqrt{|u|+|v|}}{\sqrt{u}+\sqrt{v}} \sqrt{|u-v|} \\
&=\sqrt{|u-v|}
\end{aligned} .\]
Note that one can justify that inequality
\[\frac{\sqrt{|u|+|v|}}{\sqrt{u}+\sqrt{v}} \leq 1\]
by squaring both sides since they are both positive. Thus, (3.9) is satisfied.
While every uniformly continuous function on a set \(D\) is also continuous at each point of \(D\), the converse is not true in general. The following example illustrates this point.
Let \(f:(0,1) \rightarrow \mathbb{R}\) be given by
\[f(x)=\frac{1}{x} .\]
Figure \(3.5\): Continuous but not uniformly continuous on \((0, \infty)\).
Solution
We already know that this function is continuous at every \(\bar{x} \in(0,1)\). We will show that \(f\) is not uniformly continuous on \((0,1)\). Let \(\varepsilon = 2\) and \(\delta > 0\). Set \(\delta_{0}=\min \{\delta / 2,1 / 4\}\), \(x=\delta_{0}\), and \(y=2 \delta_{0}\). Then \(x,y \in (0,1)\) and \(|x-y|=\delta_{0}<\delta\), but
\[|f(x)-f(y)|=\left|\frac{1}{x}-\frac{1}{y}\right|=\left|\frac{y-x}{x y}\right|=\left|\frac{\delta_{0}}{2 \delta_{0}^{2}}\right|=\left|\frac{1}{2 \delta_{0}}\right| \geq 2=\varepsilon .\]
This shows \(f\) is not uniformly continuous on \((0,1)\).
The following theorem offers a sequential characterization of uniform continuity analogous to that in Theorem 3.3.3.
Let \(D\) be a nonempty subset of \(\mathbb{R}\) and \(f: D \rightarrow \mathbb{R}\). Then \(f\) is uniformly continuous on \(D\) if and only if the following condition holds
(C) for every two sequences \(\left\{u_{n}\right\}\), \(\left\{v_{n}\right\}\) in \(D\) such that \(\lim _{n \rightarrow \infty}\left(u_{n}-v_{n}\right)=0\), it follows that \(\lim _{n \rightarrow \infty}\left(f\left(u_{n}\right)-f\left(v_{n}\right)\right)=0\).
- Proof
-
Suppose first that \(f\) is uniformly continuous and let \(\left\{u_{n}\right\}\), \(\left\{v_{n}\right\}\) be sequences in \(D\) such that \(\lim _{n \rightarrow \infty}\left(u_{n}-v_{n}\right)=0\). Let \(\varepsilon > 0\). Choose \(\delta > 0\) such that \(|f(u)-f(v)|<\varepsilon\) whenever \(u,v \in D\) and \(|u-v|<\delta\). Let \(N \in \mathbb{N}\) be such that \(\left|u_{n}-v_{n}\right|<\delta\) for \(n \geq N\). For such \(n\), we have \(\left|f\left(u_{n}\right)-f\left(v_{n}\right)\right|<\varepsilon\). This shows \(\lim _{n \rightarrow \infty}\left(f\left(u_{n}\right)-f\left(v_{n}\right)\right)=0\).
To prove the converse, assume condition (C) holds and suppose, by way of contradiction, that \(f\) is not uniformly continuous. Then there exists \(\varepsilon_{0}>0\) such that for any \(\delta > 0\), there exists \(u,v \in D\) with
\[|u-v|<\delta \text { and }|f(u)-f(v)| \geq \varepsilon_{0} .\]
Thus, for every \(n \in \mathbb{N}\), there exist \(u_{n}, v_{n} \in D\) with
\[\left|u_{n}-v_{n}\right| \leq 1 / n \text { and }\left|f\left(u_{n}\right)-f\left(v_{n}\right)\right| \geq \varepsilon_{0} .\]
It follows that for such sequences, \(\lim _{n \rightarrow \infty}\left(u_{n}-v_{n}\right)=0\), but \(\left\{f\left(u_{n}\right)-f\left(v_{n}\right)\right\}\) does not converge to zero, which contradicts the assumption. \(\square\)
Using this theorem, we can give an easier proof that the function in Example 3.5.6 is not uniformly continuous.
Solution
Consider the two sequences \(u_{n}=1 /(n+1)\) and \(v_{n}=1 / n\) for all \(n \geq 2\). Then clearly, \(\lim _{n \rightarrow \infty}\left(u_{n}-v_{n}\right)=0\), but
\[\lim _{n \rightarrow \infty}\left(f\left(u_{n}\right)-f\left(v_{n}\right)\right)=\lim _{n \rightarrow \infty}\left(\frac{1}{1 /(n+1)}-\frac{1}{1 / n}\right)=\lim _{n \rightarrow \infty}(n+1-n)=1 \neq 0 .\]
The following theorem shows one important case in which continuity implies uniform continuity.
Let \(f: D \rightarrow \mathbb{R}\) be a continuous function. Suppose \(D\) is compact. Then \(f\) is uniformly continuous on \(D\).
- Proof
-
Suppose by contradition that \(f\) is not uniformly continuous on \(D\). Then there exists \(\varepsilon_{0}>0\) such that for any \(\delta > 0\), there exists \(u,v \in D\) with
\[|u-v|<\delta \text { and }|f(u)-f(v)| \geq \varepsilon_{0} .\]
Thus, for every \(n \in \mathbb{N}\), there exists \(u_{n}, v_{n} \in D\) with
\[\left|u_{n}-v_{n}\right| \leq 1 / n \text { and }\left|f\left(u_{n}\right)-f\left(v_{n}\right)\right| \geq \varepsilon_{0} .\]
Since \(D\) is compact, there exist \(u_{0} \in D\) and a subsequence \(\left\{u_{n_{k}}\right\}\) of \(\left\{u_{n}\right\}\) such that
\[u_{n_{k}} \rightarrow u_{0} \text { as } k \rightarrow \infty .\]
Then
\[\left|u_{n_{k}}-v_{n_{k}}\right| \leq \frac{1}{n_{k}} .\]
for all \(k\) and, hence, we also have
\[v_{n_{k}} \rightarrow u_{0} \text { as } k \rightarrow \infty .\]
By the continuity of \(f\),
\[f\left(u_{n_{k}}\right) \rightarrow f\left(u_{0}\right) \text { and } f\left(v_{n_{k}}\right) \rightarrow f\left(u_{0}\right) .\]
Therefore, \(f\) converges to zero, which is a contradiction. The proof is now complete.
\(\square\)
We now prove a result that characterizes uniform continuity on open bounded intervals. We first make the observation that if \(f: D \rightarrow \mathbb{R}\) is uniformly continuous on \(D\) and \(A \subset D\), then \(f\) is uniformly continuous on \(A\). More precisely, the restriction \(f_{\mid A}: A \rightarrow \mathbb{R}\) is uniformly continuous on \(A\) (see Section 1.2 for the notation). This follows by noting that if \(|f(u)-f(v)|<\varepsilon\) whenever \(u,v \in D\) with \(|u-v|<\delta\), then we also have \(|f(u)-f(v)|<\varepsilon\) when we restrict \(u,v\) to be in \(A\).
Let \(a,b \in \mathbb{R}\) and \(a < b\). A function \(f:(a, b) \rightarrow \mathbb{R}\) is uniformly continuous if and only if \(f\) can be extended to a continuous function \(\tilde{f}:[a, b] \rightarrow \mathbb{R}\) (that is, there is a continuous function \(\tilde{f}:[a, b] \rightarrow \mathbb{R}\) such that \(f=\tilde{f}_{\mid(a, b)}\)).
- Proof
-
Suppose first that there exists a continuous function \(\tilde{f}:[a, b] \rightarrow \mathbb{R}\) such that \(f=\tilde{f}_{\mid(a, b)}\). By Theorem 3.5.4, the function \(\tilde{f}\) is uniformly continuous on \([a,b]\). Therefore, it follows from our early observation that \(f\) is uniformly continuous on \((a,b)\).
For the converse, suppose \(f:(a, b) \rightarrow \mathbb{R}\) is uniformly continuous. We will show first that \(\lim _{x \rightarrow a^{+}} f(x)\) exists. Note that the one sided limit corresponds to the limit in Theorem 3.2.2. We will check that the \(\varepsilon-\delta\) condition of Theorem 3.2.2 holds.
Let \(\varepsilon > 0\). Choose \(\delta_{0}>0\) so that \(|f(u)-f(v)|<\varepsilon\) whenever \(u,v \in (a,b)\) and \(|u-v|<\delta_{0}\). Set \(\delta=\delta_{0} / 2\). Then, if \(u, v \in(a, b)\), \(|u-a|<\delta\), and \(|v-a|<\delta\) we have
\[|u-v| \leq|u-a|+|a-v|<\delta+\delta=\delta_{0}\]
and, hence, \(|f(u)-f(v)|<\varepsilon\). We can now invoke Theorem 3.2.2 to conclude \(\lim _{x \rightarrow a^{+}} f(x)\) exists. In a similar way we can show that \(\lim _{x \rightarrow b^{-}} f(x)\) exists. Now define, \(\tilde{f}:[a, b] \rightarrow \mathbb{R}\) by
\[\tilde{f}(x)=\left\{\begin{array}{ll}
f(x), & \text { if } x \in(a, b) \text {;} \\
\lim _{x \rightarrow a^{+}} f(x), & \text { if } x=a \text {;} \\
\lim _{x \rightarrow b^{-}} f(x), & \text { if } x=b \text {.}
\end{array}\right .\]By its definition \(\tilde{f}_{\mid(a, b)}=f\) and, so, \(tilde{f}\) is continuous at every \(x \in (a,b)\). Moreover, \(\lim _{x \rightarrow a^{+}} \tilde{f}(x)= \lim _{x \rightarrow a^{+}} f(x)=\tilde{f}(a)\) and \(\lim _{x \rightarrow b^{-}} \tilde{f}(x)=\lim _{x \rightarrow b^{-}} f(x)=\tilde{f}(b)\), so \(\tilde{f}\) is also continuous at \(a\) and \(b\) by Theorem 3.3.2. Thus \(\tilde{f}\) is the desired continuous extension of \(f\). \(\square\)
Exercise \(\PageIndex{1}\)
Prove that each of the following functions is uniformly continuous on the given domain:
- \(f(x)=a x+b, a, b \in \mathbb{R}\), on \(\mathbb{R}\).
- \(f(x)=1 / x \text { on }[a, \infty)\), where \(a > 0\).
- Answer
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Exercise \(\PageIndex{2}\)
Prove that each of the following functions is not uniformly continuous on the given domain:
- \(f(x)=x^{2}\) on \(\mathbb{R}\).
- \(f(x)=\sin \frac{1}{x}\) on \((0,1)\).
- \(f(x)=\ln (x)\) on \((0, \infty)\).
- Answer
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Exercise \(\PageIndex{3}\)
Determine which of the following functions are uniformly continuous on the given domains.
- \(f(x)=x \sin \left(\frac{1}{x}\right)\) on \((0,1)\).
- \(f(x)=\frac{x}{x+1}\) on \([0, \infty)\).
- \(f(x)=\frac{1}{|x-1|}\) on \((0,1)\).
- \(f(x)=\frac{1}{|x-2|}\) on \((0,1)\).
- Answer
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Exercise \(\PageIndex{4}\)
Let \(D \subset \mathbb{R}\) and \(k \in \mathbb{R}\). Prove that if \(f, g: D \rightarrow \mathbb{R}\) are uniformly continuous on \(D\), then \(f+g\) and \(kf\) are uniformly continuous on \(D\).
- Answer
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Exercise \(\PageIndex{5}\)
Give an example of a subset \(D\) of \(\mathbb{R}\) and uniformly continuous functions \(f, g: D \rightarrow \mathbb{R}\) such that \(fg\) is not uniformly continuous on \(D\).
- Answer
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Exercise \(\PageIndex{6}\)
Let \(D\) be a nonempty subset of \(\mathbb{R}\) and let \(f: D \rightarrow \mathbb{R}\). Suppose that \(f\) is uniformly continuous on \(D\). Prove that if \(\left\{x_{n}\right\}\) is a cauchy sequence with \(x_{n} \in D\) for every \(n \in \mathbb{N}\), then \(\left\{f\left(x_{n}\right)\right\}\) is also a Caucy sequence.
- Answer
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Exercise \(\PageIndex{7}\)
Let \(a,b \in \mathbb{R}\) and let \(f:(a, b) \rightarrow \mathbb{R}\).
- Prove that if \(f\) is uniformly continuous, then \(f\) is bounded.
- Prove that if \(f\) is continuous, bounded, and monotone, then it is uniformly continuous.
- Answer
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Exercise \(\PageIndex{8}\)
Let \(f\) be a continuous function on \([a, \infty)\). Suppose
\[\lim _{x \rightarrow \infty} f(x)=c .\]
- Prove that \(f\) is bounded on \([a, \infty)\).
- Prove that \(f\) is uniformly continuous on \([a, \infty)\).
- Suppose further that \(c > f(a)\). Prove that there exists \(x_{0} \in[a, \infty)\) such that
\[f\left(x_{0}\right)=\inf \{f(x): x \in[a, \infty)\}.\]
- Answer
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Source: https://math.libretexts.org/Bookshelves/Analysis/Introduction_to_Mathematical_Analysis_I_(Lafferriere_Lafferriere_and_Nguyen)/03%3A_Limits_and_Continuity/3.05%3A_Uniform_Continuity
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