Proof. Let \(f\) be uniformly continuous on \(I\). Then for each \(\varepsilon > 0\, \exists \, \delta > 0 \ni \) for \(x\, ,y\in I\).
\(\begin {vmatrix} f(x) - f(y)\\ \end {vmatrix} < \varepsilon \) whenever \(\begin {vmatrix} x - y\\ \end {vmatrix} < \delta \). Let \(p \in I\) be any point. Since \(I\) is an interval, every sequence in \(I\) converging to \(p\) is either monotone increasing or monotone decreasing. Let \(\, \big \{x_n\big \}^{\infty }_{n = 1}\) be any monotone sequence in \(I\) such that \(\displaystyle {\lim _{n\rightarrow \infty } x_n = p}\). Then for each \(\delta > 0\, \exists \, N\in \mathbb {N} \ni \begin {vmatrix} x_n - p\\ \end {vmatrix} < \delta \, ,\, \forall \, n >N\). \(\implies \begin {vmatrix} f(x) - f(p)\\ \end {vmatrix} < \varepsilon \,\,\forall \, n> N\implies \displaystyle {\lim _{n\rightarrow \infty } f(x_n) = f(p)}\). Therefore \(f\) is continuous at \(p\). Since \(p\) is an arbitrary point in \(I\), \(f\) is continuous on \(I\).

However, a continuous function is not necessary uniformly continuous. For example, a function \(f\) defined by \(f(x) = x^2\) is continuous on \(\mathbb {R}\), but not uniformly continuous since for \(\delta > 0\) there is a positive integer \(N\) such that \(\dfrac {1}{N}<\delta \). Take \(x = N\) and \(y = N + \dfrac {1}{N}\). Clearly \(x\, , x\in \mathbb {R}\). Now, \(\begin {vmatrix} x - y\\ \end {vmatrix} = \dfrac {1}{N} < \delta \). But \(\begin {vmatrix} f(x) - f(y)\\ \end {vmatrix} = \begin {vmatrix} N^2 - \Big (N + \dfrac {1}{N}\Big )^2\\ \end {vmatrix} = 2 + \dfrac {1}{N^2}> 2 = \varepsilon \). By definition, \(f\) is not uniformly continuous.

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Proof. Suppose \(f\) is not uniformly continuous on \([a,b]\). Then there exist an \(\varepsilon _0 > 0\) for which
\(\forall \, \delta \big (= 1/n\big )>0\, n\in \mathbb {N}\) there are points \(x_n,\, y_n\in [a,b]\ni \begin {vmatrix} x_n - y_n\\ \end {vmatrix} < 1/n\) and
\(\begin {vmatrix} f(x_n) - f(y_n)\\ \end {vmatrix}\geq \varepsilon _0\hspace {0.3cm}\cdots \cdots \hspace {0.3cm} (*)\). Thus, we have sequences \(\big \{x_n\big \}^{\infty }_{n = 1}\) is and \(\big \{y_n\big \}^{\infty }_{n =1}\) which satisfy \((*)\). But \(\big \{x_n\big \}^{\infty }_{n =1}\) is a bounded sequence, hence it has a convergent subsequence \(\big \{x_{n_k}\big \}^{\infty }_{k = 1}\). Let \(\displaystyle {\lim _{k\rightarrow \infty }\, x_{n_k} = x}\). Then \(x \in [a,b]\) since \([a,b]\) is closed. Let \(\big \{y_{n_k}\big \}^{\infty }_{k = 1}\) be a subsequence for \(\big \{y_n\big \}\). Then form \((*)\) we have \(\begin {vmatrix} x_{n_k} - y_{n_k}\\ \end {vmatrix} < \dfrac {1}{n_k}\) and \(\begin {vmatrix} f(x_{n_k})- f(y_{n_k})\\ \end {vmatrix}\geq \varepsilon _0\hspace {0.3cm}\cdots \cdots \hspace {0.3cm}(**)\). Note that \begin {align*} \begin {vmatrix} y_{n_k} - x\\ \end {vmatrix} & \leq \begin {vmatrix} y_{n_k} - x_{n_k}\\ \end {vmatrix} + \begin {vmatrix} x_{n_k} - x\\ \end {vmatrix}\\ & < \dfrac {1}{n_k} + \begin {vmatrix} x_{n_k} - x\\ \end {vmatrix}\longrightarrow 0 \end {align*}

as \(k\longrightarrow \infty \), since \(\displaystyle {\lim _{k\rightarrow \infty }\, x_{n_k} = x}\). Thus \(\displaystyle {\lim _{k\rightarrow \infty }\, y_{n_k} = x}\). Now, since \(f\) is continuous at \(x\) and \(\displaystyle {\lim _{k\rightarrow \infty }\, x_{n_k} = x}\) \(\, , \, \displaystyle {\lim _{k\rightarrow \infty }\, f(x_{n_k}) = f(x)}\). Thus, for \(\varepsilon > 0\, \exists \, N\in \mathbb {N} \ni \begin {vmatrix} f(x_{n_k}) - f(x)\\ \end {vmatrix} < \varepsilon \hspace {0.2cm} \forall \, n> N\). Therefore \begin {align*} \begin {vmatrix} f(y_{n_k}) - f(x)\\ \end {vmatrix} & = \begin {vmatrix} f(y_{n_k}) - f(x_{n_k}) + f(x_{n_k}) - f(x)\\ \end {vmatrix}\\ & = \begin {vmatrix} f(y_{n_k}) - f(x_{n_k}) - \big (f(x) - f(x_{n_k})\big )\\ \end {vmatrix}\\ & \geq \begin {vmatrix} f(y_{n_k}) - f(x_{n_k})\\ \end {vmatrix} - \begin {vmatrix} f(x) - f(x_{n_k})\\ \end {vmatrix}\\ & \geq \varepsilon _0 - \varepsilon \end {align*}

Thus \(\big \{y_{n_k}\big \}^{\infty }_{k = 1}\) does not converge to \(x\). Contradicting that \(f\) is continuous at \(x\in [a,b]\). Therefore, \(f\) must be uniformly continuous.

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Proof. Uniform continuity implies continuity trivially, by fixing \(x\) in the definition. For the converse, suppose \(f\) is continuous on \([a,b]\) but not uniformly continuous. Then some \(\varepsilon _0>0\) admits no \(\delta \), so for each \(n\) there are \(x_n,y_n\in [a,b]\) with \[\left |x_n-y_n\right | < \frac 1n \qquad \text {but}\qquad \left |f(x_n)-f(y_n)\right | \geq \varepsilon _0 .\] The sequence \((x_n)\) lies in a closed bounded interval, so by Bolzano–Weierstrass it has a subsequence \(x_{n_k}\rightarrow x\in [a,b]\). Since \(\left |x_{n_k}-y_{n_k}\right |\rightarrow 0\), also \(y_{n_k}\rightarrow x\). Continuity at \(x\) then forces both \(f(x_{n_k})\) and \(f(y_{n_k})\) to converge to \(f(x)\), so their difference tends to \(0\) — contradicting that it stays at least \(\varepsilon _0\). □

Proof. Since \(f\) is uniformly continuous on \(I\), for each \(\varepsilon > 0\, \exists \, \delta > 0 \ni \, \forall \, x\, ,\,y\in I,\hspace {0.2cm}\begin {vmatrix} f(x) - f(y)\\ \end {vmatrix} < \varepsilon \) whenever \(\begin {vmatrix} x - y\\ \end {vmatrix} < \delta \).
Also, since \(\big \{x_n\big \}^{\infty }_{n = 1}\) is a Cauchy sequence, for \(\delta > 0\, \exists \, N\in \mathbb {N}\ni \begin {vmatrix} x_n - x_m\\ \end {vmatrix} < \delta \hspace {0.2cm} \forall \, m,\, n > N\).
Thus, for each \(\varepsilon > 0\, \exists \, N\in \mathbb {N} \ni \begin {vmatrix} f(x_n) - f(x_m)\\ \end {vmatrix} < \varepsilon \hspace {0.2cm} \forall \, m\, , n> N\). Hence, \(\big \{f(x_n)\big \}^{\infty }_{n = 1}\) is a Cauchy sequence.

However, the above theorem does not hold for mere continuous functions. For example, we consider the function, \(f(x) = \dfrac {1}{x}\, ,\, x\in (0,1]\), which is not uniformly continuous on \((0,1]\).
Taking \(x_n = \dfrac {1}{n}\, ,\, n \in \mathbb {N}\), we have a Cauchy sequence \(\big \{x_n\big \}\). But \(\big \{x_n\big \}\) is not a Cauchy sequence since \(f(x_n) = n\), which diverges.

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Proof. Let \(K\) be a Lipschitz constant, so \(\left |f(x)-f(y)\right |\leq K\left |x-y\right |\) for all \(x,y\in A\). If \(K=0\) then \(f\) is constant and the result is trivial. Otherwise, given \(\varepsilon >0\) put \(\delta = \varepsilon /K\). Then \(\left |x-y\right |<\delta \) gives \[\left |f(x)-f(y)\right | \leq K\left |x-y\right | < K\cdot \frac {\varepsilon }{K} = \varepsilon .\] The \(\delta \) depends on \(\varepsilon \) alone, which is uniform continuity. □

Proof. For each \(x\in K\) and \(\varepsilon > 0\) there exists \(\delta (\varepsilon , x) > 0\) such that if \(\begin {vmatrix} x - y\\ \end {vmatrix} < \delta (\varepsilon , x)\) then \(\begin {vmatrix} f(x) - f(y)\\ \end {vmatrix} < \dfrac {\varepsilon }{2}\). Now for each \(x\in K\) let \(I(x) = \big \{y:\, \begin {vmatrix} y - x\\ \end {vmatrix} < \delta (\varepsilon , x)/2\big \}\). Then the collection \(C = \big \{I(x):\, x\in K\big \}\) is an open cover for the compact set \(K\) and hence there are finitely many points \(x_1,\, x_2,\, \cdots \cdots \, x_n\) in \(K\) such that \(K\subset I(x_1) \cup I(x_2)\cup \cdots \cdots \cup I(x_n)\).
Let \(\delta = \min \big \{\delta (\varepsilon , x_p)/2:\, p = 1,\, 2,\, \cdots \cdots \, , n\big \}\). Suppose now that \(x,\, y\in K\) and \(\begin {vmatrix} x - y\\ \end {vmatrix} < \delta \). Then \(x\) belongs to some \(I(x_p)\). This implies that \begin {align*} \begin {vmatrix} y - x_p\\ \end {vmatrix} & = \begin {vmatrix} y - x + x - x_p\\ \end {vmatrix}\\ & \leq \begin {vmatrix} y - x\\ \end {vmatrix} + \begin {vmatrix} x - x_p\\ \end {vmatrix}\\ & < \delta + \delta (\varepsilon , x_p)/2\\ & < \delta (\varepsilon , x_p) \end {align*}

Hence \(\begin {vmatrix} f(y) - f(x)\\ \end {vmatrix}\leq \begin {vmatrix} f(y) - f(x_p)\\ \end {vmatrix} + \begin {vmatrix} f(x_p) - f(x)\\ \end {vmatrix} < \varepsilon \) which proves that \(f\) is uniformly continuous on \(K\).

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