5. Continuity

Let \(c\) be irrational and let \(\eps=1/2\). Then for all \(\delta \gt 0\), there exists \(x\in\mathbb{Q}\cap (c-\delta, c+\delta)\) (by the Density theorem) and therefore \(|f(x)-f(c)|=1 \gt \eps\). Hence, \(f\) is discontinuous at \(c\). A similar argument shows that \(f\) is discontinuous \(c\in\mathbb{Q}\). Alternatively, if \(c\in\mathbb{Q}\) then there exists a sequence of irrational numbers \((x_n)\) converging to \(c\). Now \(f(x_n)=0\) and \(f(c)=1\), and this proves that \(f\) is discontinuous at \(c\). A similar arguments holds for \(c\) irrational.

Let \(c=\frac{m}{n}\in\mathbb{Q}\) with \(\gcd(m,n)=1\). There exists a sequence \((x_n)\) of irrational numbers in \(A\) converging to \(c\). Hence, \(f(x_n)=0\) while \(f(c)=\frac{1}{n}\). This shows that \(f\) is discontinuous at \(c\). Now let \(c\) be irrational and let \(\eps \gt 0\) be arbitrary. Let \(N\in\N\) be such that \(\frac{1}{N} \lt \eps\). In the interval \((c-1,c+1)\), there are only a finite number of rationals \(\frac{m}{n}\) with \(n \lt N\), otherwise we can create a sequence \(\frac{m_k}{n_k}\) with \(n_k \lt N\), all the rationals \(\frac{m_k}{n_k}\) distinct and thus necessarily \(\frac{m_k}{n_k}\) is unbounded. Hence, there exists \(\delta \gt 0\) such that the interval \((c-\delta,c+\delta)\) contains only rational numbers \(x=\frac{m}{n}\) with \(n \gt N\). Hence, if \(x=\frac{m}{n}\in (c-\delta,c+\delta)\) then \(f(x)=\frac{1}{n} \lt \frac{1}{N}\) and therefore \(|f(x)-f(c)|=\frac{1}{n} \lt \frac{1}{N} \lt \eps\). On the other hand, if \(x\in(c-\delta,c+\delta)\) is irrational then \(|f(x)-f(c)|=|0-0| \lt \eps\). This proves that \(f\) is continuous at \(c\).

Let \(\eps \gt 0\) be arbitrary. By continuity of \(f\) and \(g\) at \(c\), there exists \(\delta_1 \gt 0\) such that \(|f(x)-f(c)| \lt \eps/2\) for all \(x\in A\) such that \(0 \lt |x-c| \lt \delta_1\), and there exists \(\delta_2 \gt 0\) such that \(|g(x)-g(c)| \lt \eps/2\) for all \(x\in A\) such that \(0 \lt |x-c| \lt \delta_2\). Let \(\delta=\min\{\delta_1,\delta_2\}\). Then for \(x\in A\) such that \(0 \lt |x-c| \lt \delta\) we have that \begin{align*} |f(x)+g(x)-(f(c)+g(c))| &\leq |f(x)-f(c)| + |g(x)-g(c)|\\ & \lt \eps/2+\eps/2\\ & = \eps. \end{align*} This proves that \(f+g\) is continuous at \(c\). A similar proof holds for \(f-g\). Consider now the function \(bf\). If \(b=0\) then \(bf(x)=0\) for all \(x\in A\) and continuity is trivial. So assume that \(b\neq 0\). Let \(\eps \gt 0\) be arbitrary. Then there exists \(\delta \gt 0\) such that if \(x\in A\cap (c-\delta,c+\delta)\), \(x\neq c\), then \(|f(x)-f(c)| \lt \eps/(|b|)\). Therefore, for \(x\in A\cap (c-\delta,c+\delta)\), \(x\neq c\), we have that \begin{align*} |bf(x)-bf(c)| &= |b| |f(x)-f(c)| \\ & \lt |b| \eps/(|b|)\\ & = \eps. \end{align*} We now prove continuity of \(fg\). Let \((x_n)\) be any sequence in \(A\) converging to \(c\). Then \(y_n=f(x_n)\) converges to \(f(c)\) by continuity of \(f\) at \(c\), and \(z_n=g(x_n)\) converges to \(g(c)\) by continuity of \(g\) at \(c\). Hence the sequence \(w_n=y_nz_n\) converges to \(f(c)g(c)\). Hence, for every sequence \((x_n)\) converging to \(c\), \(f(x_n)g(x_n)\) converges to \(f(c)g(c)\). This shows that \(fg\) is continuous at \(c\).

Let \(\eps \gt 0\) be arbitrary. Let \(\delta=\eps\). If \(0 \lt |x-c| \lt \delta\) then \(|f(x)-f(c)|=|x-c| \lt \delta=\eps\).

Let \(\eps \gt 0\) be arbitrary. Then there exists \(\delta \gt 0\) such that \(|f(y)-f(d)| \lt \eps\) for all \(0 \lt |y-d| \lt \delta\). Therefore, if \(0 \lt |x-c|=|(x+\alpha)-(c+\alpha)| \lt \delta\) then \(|f(x+\alpha)-f(c+\alpha)| \lt \eps\) and therefore \[ |g(x)-g(c)| = |f(x+\alpha)-f(c+\alpha)| \lt \eps. \]

We have that \begin{align*} |\sin(x)-\sin(c)| & \leq 2|\sin(\tfrac{1}{2}(x-c))| |\cos(\tfrac{1}{2}(x-c))| \\[2ex] &\leq 2 \frac{1}{2} |x-c| \\[2ex] &= |x-c|. \end{align*} Hence given \(\eps \gt 0\) we choose \(\delta=\eps\). The proof that \(\cos(x)\) is continuous follows from the fact that \(\cos(x)=\sin(x+\pi/2)\) and Lemma 5.2.6.

For \(c=0\), we must consider \(|\sqrt{x}-\sqrt{0}| = \sqrt{x}\). Given \(\eps \gt 0\) let \(\delta=\eps^2\). Then if \(x\in A\) and \(x \lt \delta=\eps^2\) then \(\sqrt{x} \lt \eps\). This shows that \(f\) is continuous at \(c=0\). Now suppose that \(c\neq 0\). Then \begin{align*} |\sqrt{x}-\sqrt{c}| &= |\sqrt{x}-\sqrt{c}| \cdot \frac{\sqrt{x}+\sqrt{c}}{\sqrt{x}+\sqrt{c}}\\[2ex] & = \frac{|x-c|}{\sqrt{x}+\sqrt{c}}\\[2ex] &\leq \frac{1}{\sqrt{c}} |x-c|. \end{align*} Hence, given \(\eps \gt 0\), suppose that \(0 \lt |x-c| \lt \sqrt{c}\eps\). Then \(|\sqrt{x}-\sqrt{c}| \lt \eps\).

Follows from the inequality \(||x|-|c||\leq |x-c|\).

Let \(\eps \gt 0\) be given. Let \(c\in A\) and let \(d=f(c)\in B\). Then there exists \(\delta_1 \gt 0\) such that if \(0 \lt |y-d| \lt \delta_1\) then \(|g(y)-g(d)| \lt \eps\). Now since \(f\) is continuous at \(c\), there exists \(\delta_2 \gt 0\) such that if \(0 \lt |x-c| \lt \delta_2\) then \(|f(x)-f(c)| \lt \delta_1\). Therefore, if \(0 \lt |x-c| \lt \delta_2\) then \(|f(x)-d| \lt \delta_1\) and therefore \(|g(f(x))-g(d)| \lt \eps\). This proves that \((g\circ f)\) is continuous at \(c\in A\). Since \(c\) is arbitrary, \((g\circ f)\) is continuous on \(A\).

Suppose that \(f\) is unbounded. Then for each \(n\in\mathbb{N}\) there exists \(x_n\in [a,b]\) such that \(|f(x_n)| \gt n\). Now the sequence \((x_n)\) is bounded since \(a\leq x_n\leq b\). By the Bolzano-Weierstrass theorem, \((x_n)\) has a convergent subsequence, say \((x_{n_k})\), whose limit \(u=\lim_{k\rightarrow\infty}x_{n_k}\) satisfies \(a\leq u\leq b\). Since \(f\) is continuous at \(u\) then \(\lim_{k\rightarrow\infty} f(x_{n_k})\) exists and equal to \(f(u)\). This is a contradiction since \(|f(x_{n_k})| \gt n_k \geq k\) implies that \(f(x_{n_k})\) is unbounded.

Let \(S=\{f(x)\;|\; x\in [a,b]\}\) be the range of \(f\). By Theorem 5.3.3, \(\sup(S)\) exists; set \(M=\sup(S)\). By the definition of the supremum, for each \(\eps \gt 0\) there exists \(x\in [a,b]\) such that \(M-\eps \lt f(x)\leq M\). In particular, for \(\eps_n=1/n\), there exists \(x_n\in [a,b]\) such that \(M-\eps_n \lt f(x_n)\leq M\). Then \(\lim_{n\rightarrow\infty} f(x_n) = M\). The sequence \((x_n)\) is bounded and thus has a convergent subsequence, say \((x_{n_k})\). Let \(x^* = \lim_{k\rightarrow\infty} x_{n_k}\). Clearly, \(a\leq x^* \leq b\). Since \(f\) is continuous at \(x^*\), we have that \(M = \lim_{k\rightarrow\infty} f(x_{n_k}) = f(x^*)\). Hence, \(x^*\) is a maximum point. A similar proof establishes that \(f\) has a minimum point on \([a,b]\).

Let \(S=\{x\in [a,b]\;:\; f(x) \lt \xi \}\). The set \(S\) is non-empty because \(a\in S\). Moreover, \(S\) is clearly bounded above. Let \(c=\sup(S)\). By the definition of the supremum, there exists a sequence \((x_n)\) in \(S\) such that \(\limi x_n = c\). Since \(a\leq x_n\leq b\) it follows that \(a \leq c \leq b\). By definition of \(x_n\), \(f(x_n) \lt \xi\) and since \(f\) is continuous at \(c\) we have that \(f(c) = \limi f(x_n) \leq \xi\), and thus \(f(c)\leq \xi\). Now let \(\delta_n \gt 0\) be such that \(\delta_n\rightarrow 0\) and \(c+\delta_n \lt b\). Then \(z_n = c+\delta_n\) converges to \(c\) and \(\xi \leq f(z_n)\) because \(z_n\not\in S\). Therefore, since \(\limi f(z_n) = f(c)\) we have that \(\xi \leq f(c)\). Therefore, \(f(c)\leq \xi \leq f(c)\) and this proves that \(\xi=f(c)\). This shows also that \(a \lt c \lt b\).

Let \(f(t)\) be the distance traveled along the trail on the way up the mountain after \(t\) units of time, and let \(g(t)\) be the distance remaining to travel along the trail on the way down the mountain after \(t\) units of time. Both \(f\) and \(g\) are defined on the same time interval, say \([0,12]\) if \(t\) is measured in hours. If \(M\) is the length of the trail, then \(f(0)=0\), \(f(12)=M\), \(g(0)=M\) and \(g(12)=0\). Let \(h(t)=g(t)-f(t)\). Then \(h(0)=M\) and \(h(12)=-M\). Hence, there exists \(t^*\in (0,12)\) such that \(h(t^*)=0\). In other words, \(f(t^*)=g(t^*)\), and therefore \(t^*\) is the time of day when the hiker is at exactly the same point on the trail.

The function \(f\) is continuous on \([0,1]\). We have that \(f(0)=-2 \lt 0\) and \(f(1)=e-2 \gt 0\). Therefore, there exists \(x^*\in (0,1)\) such that \(f(x^*)=0\), i.e., \(f\) has a zero in the interval \((0,1)\).

Since \(f\) achieves its maximum and minimum value on \([a,b]\), there exists \(x^*,x_*\in [a,b]\) such that \(f(x_*)\leq f(x)\leq f(x^*)\) for all \(x\in [a,b]\). Hence, \(f([a,b])\subset [f(x_*),f(x^*)]\). Assume for simplicity that \(x_* \lt x^*\). Then \([x_*,x^*]\subset [a,b]\). Let \(\xi\in\real\) be such that \(f(x_*) \lt \xi \lt f(x^*)\). Then by the Intermediate Value Theorem, there exists \(c \in (x_*,x^*)\) such that \(\xi=f(c)\). Hence, \(\xi \in f([a,b])\), and this shows that \([f(x_*),f(x^*)] \subset f([a,b])\). Therefore, \(f([a,b]) = [f(x_*),f(x^*)]=[\min(f), \max(f)]\).

We have that \(|f(x)-f(c)|=|kx-kc|=|k||x-c|\). Hence, for any \(\eps \gt 0\) we let \(\delta=\eps/|k|\), and thus if \(|x-c| \lt \delta\) then \(|f(x)-f(c)| \lt \eps\).

We have that \begin{align*} |\sin(x)-\sin(c)| &\leq 2 |\sin(\tfrac{1}{2}(x-c))|\\ & \leq |x-c|. \end{align*} Hence, for \(\eps \gt 0\) let \(\delta=\eps\) and if \(|x-c| \lt \delta\) then \(|\sin(x)-\sin(c)| \lt \eps\).

We have that \begin{align*} |f(x)-f(c)| &= \left|\frac{1}{1+x^2}-\frac{1}{1+c^2}\right| \\[2ex] &= \left|\frac{1+c^2-1-x^2}{(1+x^2)(1+c^2)}\right|\\[2ex] &= \left|\frac{x+c}{(1+x^2)(1+c^2)}\right| |x-c|\\[2ex] &= \left|\frac{x}{(1+x^2)(1+c^2)}+\frac{c}{(1+x^2)(1+c^2)}\right| |x-c|. \end{align*} Now, \(|x|\leq 1+x^2\) implies that \(\frac{|x|}{1+x^2}\leq 1\) and therefore \[ \frac{|x|}{(1+x^2)(1+c^2)} \leq \frac{1}{1+c^2} \leq 1. \] It follows that \(|f(x)-f(c)|\leq 2|x-c|\). Hence, given \(\eps \gt 0\) we let \(\delta=\eps/2\), and if \(|x-c| \lt \delta\) then \(|f(x)-f(c)| \lt \eps\).

We prove the contrapositive, that is, we will prove that if \(f\) is not uniformly continuous on \([a,b]\) then \(f\) is not continuous on \([a,b]\). Suppose then that \(f\) is not uniformly continuous on \([a,b]\). Then there exists \(\eps \gt 0\) such that for \(\delta_n=1/n\), there exists \(x_n,u_n\in [a,b]\) such that \(|x_n-u_n| \lt \delta_n\) but \(|f(x_n)-f(u_n)|\geq \eps\). Clearly, \(\lim (x_n-u_n) = 0\). Now, since \(a\leq x_n\leq b\), by the Bolzano-Weierstrass theorem there is a subsequence \((x_{n_k})\) of \((x_n)\) that converges to a point \(z \in [a,b]\). Now, \[ |u_{n_k}-z| \leq |u_{n_k}-x_{n_k}| + |x_{n_k}-z| \] and therefore also \(\lim_{k\rightarrow\infty} u_{n_k} = z\). Hence, both \((x_{n_k})\) and \((u_{n_k})\) are sequences in \([a,b]\) converging to \(z\) but \(|f(x_{n_k})-f(u_{n_k})|\geq \eps\). Hence \(f(x_{n_k})\) and \(f(u_{n_k})\) do not converge to the same limit and thus \(f\) is not continuous at \(z\). This completes the proof.

Consider \(x_n=\sqrt{\pi n}\) and \(u_n=\sqrt{\pi n} + \frac{\sqrt{\pi}}{4\sqrt{n}}\). Clearly \(\lim (x_n-u_n)=0\). On the other hand \begin{align*} |f(x_n)-f(u_n)| &= \left|\sin(\pi n) - \sin(\sqrt{\pi n} +\tfrac{\sqrt{\pi}}{4\sqrt{n}})^2\right| \\[2ex] & = \left| \sin\left(\pi n + \tfrac{\pi}{2} + \tfrac{\pi}{16n}\right)\right| \\[2ex] &= \left| \cos \left(\pi n + \tfrac{\pi}{16n}\right)\right|\\[2ex] &= \left| \cos(n\pi)\cos(\tfrac{\pi}{16n}) - \sin(n\pi)\sin(\tfrac{\pi}{16n}) \right| \\[2ex] &= |(-1)^n \cos(\tfrac{\pi}{16n})|\\[2ex] &= \cos(\tfrac{\pi}{16n})\\[2ex] &\geq \cos(\tfrac{\pi}{16}) \end{align*}

By assumption, \(|f(x)-f(u)|\leq K |x-u|\) for all \(x,u \in A\) for some constant \(K \gt 0\). Let \(\eps \gt 0\) be arbitrary and let \(\delta=\eps/K\). Then if \(|x-u| \lt \delta\) then \(|f(x)-f(u)|\leq K |x-u| \lt \eps\). Hence, \(f\) is uniformly continuous on \(A\).