be a polynomial of degree \(m \in \mathbb{N}\) and \(\xi \in \mathbb{R}\). Then p is continuous at \(\xi\) (hence all polynomials are continuous on \(\mathbb{R}\)): Let \(\epsilon > 0\), then we know from example i) after Thm 29, that for all \(x \in U_1(\xi): |p(x)-p(\xi)| \le M |x-\xi|\), where \(M > 0\) is a constant. Hence, setting \(\delta := \min(1, \frac{\epsilon}{M})\), we have \(|p(x)-p(\xi)| < \epsilon\) for all \(x \in U_\delta(\xi)\).
ii) Generally, in order to show that a function \(f: D \to \mathbb{R} \ (D \subset \mathbb{R})\) is not continuous at some given point \(\xi \in D\), according to Thm 30, part i), it is sufficient to find a sequence \((x_m)_{m=1}^\infty\) in D, converging to \(\xi\), such that \(\lim_{m \to \infty} f(x_m) \neq f(\xi)\).
Let us apply this to show that the signum function \(sign: \mathbb{R} \to \mathbb{R}\) defined by
\(sign(x) = \begin{cases} -1, & x < 0 \\ 0, & x=0 \\ 1, & x> 0 \end{cases}\)
is not continuous at \(\xi = 0\). To this end, we choose \(x_m = \frac{1}{m}, m \in \mathbb{N}\), then we have
\(\lim_{m \to \infty} sign(x_m) = 1 \neq 0 = sign(0)\).
iii) Let \(r: D \to \mathbb{R}\) be a rational function, i.e., \(r(x) = \frac{p(x)}{q(x)}\), where \(p\) and \(q\) are polynomials and \(D = \{x \in \mathbb{R} : q(x) \neq 0\}\) is the "natural" domain of definition. In light of Thm 30, part ii), and example i) we conclude that \(r\) is continuous on D.
iv) The exponential function \(\exp: \mathbb{R} \to (0, \infty)\) is continuous on \(\mathbb{R}\): By Thm 26, part v), we know that \(\lim_{m \to \infty} \exp(x_m) = \exp(\xi)\), whenever \(\xi \in \mathbb{R}\) and \((x_m)_{m=1}^\infty\) is a sequence in \(\mathbb{R}\) converging to \(\xi\).
v) An important consequence of continuity is the following: Let \(f: D \to \mathbb{R} \ (D \subset \mathbb{R})\) be continuous at the point \(\xi \in D\) with \(f(\xi) \neq 0\). Then we can find \(\delta > 0\) such that \(f(x) \neq 0\) for all \(x \in U_\delta(\xi) \cap D\).
To see this, let us choose \(\epsilon := \frac{1}{2}|f(\xi)|\).
\(\implies\) (f cont. at \(\xi\)) \(\exists \delta > 0 \ \forall x \in U_\delta(\xi) \cap D : |f(x) - f(\xi)| < \epsilon=\frac{1}{2}|f(\xi)|\).
\(\implies\) (reverse triangle) \(\frac{1}{2}|f(\xi)| > |f(\xi)| - |f(x)|\), \(x \in U_\delta(\xi) \cap D\)
\(\implies |f(x)| > \frac{1}{2}|f(\xi)| > 0\), \(x \in U_\delta(\xi) \cap D\).
Theorem 31. (Intermediate Value Theorem)
Let \(f: [a, b] \to \mathbb{R}\) be a continuous function on the closed interval \([a, b] \ (a, b \in \mathbb{R})\). If y is any value between \(f(a)\) and \(f(b)\), i.e., \(y \in [\min(f(a), f(b)), \max(f(a), f(b))]\), then there exists a point \(x \in [a, b]\) such that \(y = f(x)\).
Proof. Case 1: \(f(a) = f(b) \quad \checkmark\)
Case 2: \(f(a) \neq f(b)\). W.l.o.g. we may assume that \(f(a) < f(b)\). Let y be a value in \((f(a), f(b))\).
We define \(M := \{ t \in [a, b] : f(t) \le y \}\).
\(\implies M \neq \emptyset\) (as \(a \in M\)), M is bounded below by a and bounded above by b.
\(\implies \sup M \in [a, b]\).
We define \(x := \sup M\) and show that \(f(x) = y\).
To this end, we choose a sequence \((t_m)_{m=1}^\infty\)
in M such that \(t_m \to x\).
\(\implies (\text{Thm 30, part i)}) \ f(x) = \lim_{m \to \infty} f(t_m) \le y < f(b)\)
\(\underbrace{ \ \ \ }_{\le y} \)
\(\implies x < b\)
Hence, we can choose a sequence \((x_m)_{m=1}^\infty\) in \((x, b]\) such that \(x_m \to x\).
\(\implies (\text{Thm 30, part i)}) \ f(x) = \lim_{m \to \infty} f(x_m) \ge y\)
\(\underbrace{ \ \ \ }_{\ge y} \)
\(\implies y \le f(x) \le y \implies y = f(x)\).
Example. Consider the equation \(e^x = x+2, \ x \in \mathbb{R}\).
We show that there exists a solution \(x_0 \in (1, 2)\).
To this end, we define \(f: \mathbb{R} \to \mathbb{R}, f(x) := e^x - x - 2\).
\(\implies f\) is continuous and \(f(1) = e-3 < 0\),
\(f(2) = e^2-4 = (e-2)(e+2) > 0 \quad (2 < e < 3)\).
\(\implies (\text{Thm 31}) \ \exists x_0 \in (1, 2)\) such that \(f(x_0) = 0\).
For the computation of \(x_0\) we would need numerical methods.
Theorem 32. Let \(I \subset \mathbb{R}\) be an interval and \(f: I \to \mathbb{R}\) a continuous function. Then \(f(I) = \{f(x) : x \in I\}\) is an interval.
Proof. Let \(\alpha, \beta \in f(I)\) such that \(\alpha \le \beta\).
\(\implies \exists a, b \in I\) such that \(\alpha = f(a), \beta = f(b)\).
\(\implies (\text{Thm 31}) \ [\alpha, \beta] \subset f(I) \implies f(I)\) is an interval.
Next we deal with inverse functions.
Definition. A function \(f: D \to \mathbb{R}\) is called
i) increasing if \(f(x_1) \le f(x_2)\)
ii) strictly increasing if \(f(x_1) < f(x_2)\)
iii) decreasing if \(f(x_1) \ge f(x_2)\)
iv) strictly decreasing if \(f(x_1) > f(x_2)\)
for all \(x_1, x_2 \in D\) such that \(x_1 < x_2\).
In all these cases we call f a (strictly) monotonic function.
Theorem 33. Let \(I \subset \mathbb{R}\) be an interval and let \(f: I \to \mathbb{R}\) be a continuous and strictly monotonic function. Then \(f: I \to J\), where \(J := f(I)\), is bijective and its inverse function \(f^{-1} : J \to I\) is continuous on J.
Proof. W.l.o.g. we may assume that f is strictly increasing. Then f is injective and, as \(J = f(I)\), \(f: I \to J\) is surjective.
\(\implies\) the inverse function \(f^{-1} : J \to I\) exists.
Now we want to show that \(f^{-1}\) is continuous.
Let \(y \in J\) and \((y_m)_{m=1}^\infty\) a sequence in J with \(y_m \to y\).
We set \(\xi := f^{-1}(y)\) and \(x_m := f^{-1}(y_m), m \in \mathbb{N}\).
Assume: \(x_m \not\to \xi \implies \exists \epsilon > 0\) such that
\(|x_m - \xi| \ge \epsilon\) for infinitely many indices m
\(\implies x_m \ge \xi + \epsilon\) for infinitely many m
or \(x_m \le \xi - \epsilon\) for infinitely many m
\(\implies\) (f is strictly increasing)
\(y_m = f(x_m) \ge f(\xi + \epsilon) > f(\xi) = y\) for infinitely many m
or \(y_m = f(x_m) \le f(\xi - \epsilon) < f(\xi)=y\) for infinitely many m.
\(\implies y_m \not\to y\), which is a contradiction.
Remark In the situation of Thm 33 it is easy to see that, if f is strictly increasing, then \(f^{-1}\) is strictly increasing. Likewise, if f is strictly decreasing, \(f^{-1}\) is strictly decreasing as well.
Example. We already know that \(\exp: \mathbb{R} \to (0, \infty)\) is continuous and strictly increasing. Hence, by Thm 33 we know that its inverse function \(\log: (0, \infty) \to \mathbb{R}\) exists, and that it is continuous and strictly increasing.
Next we turn to extreme values of continuous functions.
Definition. i) A function \(f: D \to \mathbb{R} \ (D \subset \mathbb{R})\) is bounded above if there exists a \(K \in \mathbb{R}\) such that for all \(x \in D: f(x) \le K\). Likewise, f is bounded below if there exists a \(K \in \mathbb{R}\) such that for all \(x \in D: f(x) \ge K\). Moreover, f is bounded if there exists a \(K \in \mathbb{R}\) such that for all \(x \in D: |f(x)| \le K\).
ii) For a function \(f: D \to \mathbb{R} \ (D \subset \mathbb{R})\) we call
\(\sup_{x \in D} f(x) := \sup f(D)\) the supremum of f (on D),
\(\inf_{x \in D} f(x) := \inf f(D)\) the infimum of f (on D).
Moreover, if they exist, we call
\(\max_{x \in D} f(x) := \max f(D)\) the maximum of f (on D),
\(\min_{x \in D} f(x) := \min f(D)\) the minimum of f (on D).
Remark. \(\max_{x \in D} f(x)\) (or \(\min f(x)\)) exists if and
only if there exists \(x_0 \in D\) such that for all \(x \in D\): \(f(x) \le f(x_0)\) (or \(f(x) \ge f(x_0)\), respectively).
Examples i) The function \(f: (0, 1) \to \mathbb{R}, f(x) = \frac{1}{x}\) is not bounded above as \(\lim_{x \to 0+} f(x) = \infty\). However, for any \(\epsilon > 0\), the function \(f: (\epsilon, 1) \to \mathbb{R}, f(x) = \frac{1}{x}\), is bounded above by \(f(\epsilon) = \frac{1}{\epsilon}\).
ii) Let \(D := (0, 1]\) and \(f: D \to \mathbb{R}, f(x) := x^2\). Then \(\max_{x \in D} f(x)\) exists and is given by \(\max_{x \in D} f(x) = f(1) = 1 = \sup_{x \in D} f(x)\). Moreover, we have \(\inf_{x \in D} f(x) = 0\) but \(\min_{x \in D} f(x)\) does not exist.
Definition. A set \(D \subset \mathbb{R}\) is called compact if every sequence in D has a convergent subsequence such that its limit is an element of D. In other words:
\(\forall\) sequence \((x_n)_{n=1}^\infty\) in D \(\exists\) subsequence \((x_{m_k})_{k=1}^\infty\) such that \(\lim_{k \to \infty} x_{m_k} \in D\).
Remark. It is not difficult to show that closed and bounded intervals \([a, b], a, b \in \mathbb{R}\), are compact. However, intervals of the forms \((a, b)\), \([a, b)\), \((a, b]\) or unbounded intervals are not compact. Moreover, it can be proved that the union of a finite number of compact sets is compact, and that the intersection of an arbitrary number of compact sets is compact.
Theorem 34. Let \(D \subset \mathbb{R}\) be a compact set and \(f: D \to \mathbb{R}\) be a continuous function. Then \(\max_{x \in D} f(x)\) and \(\min_{x \in D} f(x)\) exist, i.e., f attains its maximum and its minimum.
Proof. W.l.o.g. we only show the existence of \(\max_{x \in D} f(x)\). We set \(s := \sup_{x \in D} f(x)\), which is in \(\mathbb{R}\) or \(\infty\). According to the definition of the supremum, we can find a sequence \((x_m)_{m=1}^\infty\) in D such that \(f(x_m) \to s\) as \(m \to \infty\).
\(\implies\) (D compact) there exists a convergent subsequence \((x_{m_k})_{k=1}^\infty\) of \((x_m)_{m=1}^\infty\)
such that \(\lim_{k \to \infty} x_{m_k} = \xi\) with \(\xi \in D\).
\(\implies\) (f continuous) \(s = \lim_{m \to \infty} f(x_m) = \lim_{k \to \infty} f(x_{m_k}) = f(\xi)\)
\(\implies s \in \mathbb{R}\) and \(f(\xi) = s \ge f(x)\), \(x \in D\),
\(\implies \max_{x \in D} f(x)\) exists and is given by \(f(\xi)\).