[Calculus] 5. Functions
Definition 5.1 Let \(X,Y\), and let \(P(x,y)\) be a property pertaining to two objects \(x\in X, y\in Y\), such that for every \(x\in X\), there is exactly one \(y\in Y\) for which \(P(x,y)\) is true. We define the function \(f: X\to Y\) defined by \(P\) on the domain \(X\) and range \(Y\) to be the object which, given any input \(x\in X\), assigns an output \(f(x)\in Y\), defined to be the unique object \(f(x)\) for which \(P(x,f(x))\) is true.[1]
Definition 5.2 Let \(a\in X\) and \(b\in Y\). We say that \(\lim_{x\to a} f(x) = b\) if
\[\forall \epsilon > 0 \exists \delta > 0 \forall x\in S[0 < d(x,a) < \delta \Rarr d(f(x),b) < \epsilon]\]
Remark 5.3 If there exists \(r > 0\) such that \(B_r(a) \cap S = \varnothing\), then \(\lim_{x\to a}f(x) = b\) holds for every \(b\in Y\).
Proposition 5.4 If \(a\) is a limit point of \(S\), and \(\lim_{x\to a}f(x) = b, \lim_{x\to a} f(x) = b'\), then \(b=b'\).
Proof Suppose \(b\ne b'\), and \(d(b,b') = 3\epsilon\) where \(\epsilon\) is a positive real.
By definition, for that \(\epsilon\), there exists $\delta_1,\delta_2 >0, such that
\[ 0 < d(x, a) < \delta_1 \Rarr d(f(x), b) < \epsilon \\ 0 < d(x, a) < \delta_2 \Rarr d(f(x), b') < \epsilon\]
Let \(\delta = \min\{\delta_1,\delta_2\}\). Then \(0 < d(x,a) < \delta \Rarr d(b,b') \le d(f(x), b) + d(f(x), b') < 2\epsilon\). This is a contradiction. □
Lemma 5.5 Let \(f:X\to\R\) be a function. If \(f\) is convergent at some point, then \(f\) is bounded.
Proof Suppose that \(f\) is not bounded, i.e., \(\forall M > 0 \exists x\in X[|f(x)| > M]\).
Let \(b:=\lim_{x\to a}f(x)\).
By definition of limit, \(\forall \epsilon > 0 \exists \delta > 0[0 < d(x,a) < \delta \Rarr |f(x) - b| < \epsilon]\).
Choose sufficiently big \(\epsilon\) such that \(b + \epsilon > 0\) (since \(b\) may be less than 0), and choose \(M = 2(b+\epsilon)\). Then we have
\[ M = 2(b+\epsilon) < f(x) < b + \epsilon\]
this is a contradiction. And hence \(f\) is bounded. □
Proposition 5.6 Let \((Y,d_y) = (\R^m, d_2)\) and \(f,g: S\to Y\), and let \(\lim_{x\to a} f(x) = b\) and \(\lim_{x\to a} g(x) = c\)
- \(\lim_{x\to a}\big(f(x)\pm g(x)\big) = b \pm c\);
- If \((Y,d_y) = (\R, d_2)\), then \(\lim_{x\to a}f(x)g(x) = \big(\lim_{x\to a}f(x)\big)\big(\lim_{x\to a} g(x)\big)\);
- If \((Y,d_y) = (\R, d_2)\) and \(c\ne 0\), then there exists \(\delta > 0\) such that \(g(x)\ne 0\) for all \(x(\ne a) \in B_\delta(a) \cap S\), and \(\lim_{x\to a}\frac{f(x)}{g(x)} = \frac b c\).
Proof (for short)
Proof 1.
\[|(f(x) \pm g(x)) - (b \pm c)| = |(f(x) - b) \pm (g(x - c))| \le |f(x) - b) + |g(x) - c| < 2\epsilon\]
Proof 2.
\[\begin{align*}|f(x)g(x) - bc| &= |f(x)g(x) + f(x)c - f(x)c - bc| \\ &= |f(x)(g(x) - c) + c(f(x) -b)| \\ &le |f(x)|g(x) - c| + |c||f(x) - b| \\ & < (M + |c|)\epsilon \end{align*}\]
for some \(M > 0\) such that \(|f(x)| \le M\) for all \(x\in S\) since $f must be bounded (by Lemma 5.5).
Proof 3.
Suppose that \(\delta\) does not exist, i.e., \(\forall \delta > 0\exists x(\ne a)\in B_\delta(x)\cap S[g(x) = 0 \Rarr d(g(x), 0) = 0]\).
Let \(d(c,0) = 2\epsilon\) where \(\epsilon > 0\).
By definition, \(x(\ne a)\in B_\delta(x)\cap S \Rarr d(g(x), c) < \epsilon\). Then
\[d(c,0) = 2\epsilon \le d(g(x),c) + d(g(x),0) < \epsilon\]
Thus that \(\delta\) must exist, such that \(g(x) \ne 0\) for all \(x\) in the ball. And
\[\begin{align*}|\frac{f(x)}{g(x)} - \frac{b}{c}| &= |\frac{f(x)c - g(x)b}{g(x)c}| \\ &= |\frac{f(x)c - g(x)b}{g(x)c} + f(x)g(x) - f(x)g(x)| \\ &= |\frac{f(x)(c - g(x)) + g(x)(f(x)-b)}{g(x)c}| \\ &\le \frac{|f(x)||c - g(x)| + |g(x)||f(x)-b|}{|g(x)c|} \\ &\le |f(x)||c - g(x)| + |g(x)||f(x)-b| \\ & < (|f(x)| + |g(x)|)\epsilon \\ & \le (M_1 + M_2|)\epsilon \end{align*}\]
for some \(M_1,M_2 > 0\) since \(f,g\) are both bounded (by Lemma 5.5) □
Proposition 5.7 Let \(X,Y,Z\) be metric spaces, \(S \sube X, T \sube Y\), and \(f: S\to T, g: T\to Z\). If \(\lim_{x\to a} f(x) = b\), \(\lim_{y\to b} g(y) = c\) and \(b\notin f(S)\), then \(\lim_{x\to a} (g\circ f) (x) = c\).
Proof \[\forall \epsilon > 0 \exists \delta' > 0 \forall y \in T[d(y,b) < \delta' \Rarr d(g(y), c) < \epsilon]\]
For this \(\delta' >0\), there exists a \(\delta >0\), such that for all \(x\in S\), we have
\[d(x,a) < \delta \Rarr d(f(x), b) < \delta' \Rarr d(g(f(x)), c) < \epsilon]\] □
Remark 5.8 If '\(b\notin f(S)\)' is not guaranteed, then the conclusion may be false. Here is an example.
Let \(X = Y = \R, Z = [0, + \infty]\), \(S = \R, T = [0, +\infty]\), and let \(f(x) = x^2, g(y) = 3\sqrt{y}\), and consider \(a = -1\). Then we have \(b = \lim_{x\to -1}f(x) = 1, c = \lim_{y\to 1} g(y) = 3\).
However, let \(h(x) = (g\circ f)(x) = 3x\) from \(\R\) to \([0, +\infty]\), then \(\lim_{x\to -1}h(x) = -3\).
Definition 5.9 (continuous functions) Give a function \(f: S \to Y\), and \(a\in S\). We say that \(f\) is continuous at \(a\) if \(\lim_{x\to a} f(x) = f(a)\). Furthermore, we say \(f\) is a continuous function if it is continuous at every \(a\in S\).
Proposition 5.10 The following statements are equivalent:
- \(f: S (\sube X) \to Y\) is a continuous function
- \(\forall V\underset{\text{open}}{\sube} Y \exists U\underset{\text{open}}{\sube} X[a\in U, f(a)\in V \Rarr f(U\cap S) \sube V)]\).
- \(\forall V\underset{\text{open}}{\sube} Y [f^{-1}(V) \underset{\text{open}}{\sube}S]\).
Proof Proof 1-2
For every point \(a \in S\), and \(f(a) \in Y\).
(\(\Rarr\))
For any open set \(V\sube Y\), there exists an \(\epsilon > 0\), such that \(B_\epsilon(f(a)) \sube V\) by definition.
For this \(\epsilon\), \(\exists \delta > 0 \forall x\in S[d(x,a) < \delta \Rarr d(f(x), f(a)) < \epsilon \Rarr f(x) \in V]\).
Let \(U\) be any other open set such that \(U\cap S \sube B_\delta(a)\). Then we have \(\forall x \in U\cap S\Rarr f(x) \in V\), i.e., \(f(U \cap S) \sube V\).
(\(\Larr\))
For every \(\epsilon > 0\), we can always find an open set \(V\) such that \(V \sube B_\epsilon(f(a))\).
There exist \(\delta_1 > 0\) such that \(B_{\delta_1}(a) \sube U\). And by definition, \(\exists \delta_2 > 0 [B_{\delta_2}(a) \sube S]\). Let \(\delta = \min\{\delta_1,\delta_2\}\), then \(B_\delta(a) \sube B\cap S\).
\[\forall x\in B_\delta(a) \sube U\cap S \Rarr f(x) \in V \sube B_\epsilon(f(a)) \Rarr d(f(x), f(a)) < \epsilon\]
i.e., \(\lim{x\to a} f(x) = f(a)\).
Proof 1-3
(\(\Rarr\))
\(\forall x \in f^{-1}(V) (f(x) \in V) \exists \epsilon > 0 [B_\epsilon(f(x)) \sube V]\).
Since \(f\) is continuous, \(exists \delta >0 [y\in B_\delta(x) \Rarr f(y) \in B_\epsilon(f(x)) \sube V]\), i.e., \(B_\delta(x) \sube f^{-1}(V)\).
Thus, \(f^{-1}(V)\) is open in \(S\).
(\(\Larr\))
For contradiction, suppose that \(f\) is not continuous, i.e, \(\exists x_0 \exists \epsilon > 0 \forall \delta > 0 \exists x \in S[d(x,x_0) < \delta \Rarr d(f(x),f(x_0)) \ge \epsilon]\).
Let \(V := B_\epsilon(f(x_0))(\sube Y)\). Then \(f^{-1}(V)\) is open too.
For \(x_0 \in f^{-1}(V)\), there exists an \(r > 0\), such that \(B_r(x_0) \sube f^{-1}(V)\), i.e.,
\(\forall x \in S[d(x, x_0) < r \Rarr f(x) \in V \Lrarr d(f(x),f(x_0)) < \epsilon]\). A contradiction against the supposition.
Thus \(f\) is continuous □
Proposition 5.11 Let \(f: X\to Y, g: Y \to Z\) be functions. If \(f\) is continuous at \(a\) and \(g\) is continuous at \(f(a)\), then \(g\circ f\) is continuous at \(a\).
Proof Similar as Proposition 5.7 □
Lemma 5.12 If \(f: X \to Y\), and \(A\sube B \sube Y\), then \(f^{-1}(A) \sube f^{-1}(B)\).
Proof Were this false, \(\exists x\in f^{-1}(A)[x\notin f^{-1}(B) \Lrarr f(x) \notin B]\).
However, \(f(x)\sube A\sube B\), which makes a contradiction. □
Proposition 5.13 If \(f:X\to Y\) is continuous, \(K(\sube X)\) is compact, then \(f(K)(\sube Y)\) is also compact.
Proof For any given open cover \(V_\alpha(\alpha\in A)\) of \(f(K)\), i.e., \(f(K)\sube \bigcup_{\alpha\in A}V_\alpha}\).
By Lemma 5.12, \(f^{-1}(f(K)) \sube f^{-1}(\bigcup_{\alpha\in A} V_\alpha)\). And \(K\sube f^{-1}(f(K)), f^{-1}(\bigcup_{\alpha\in A} V_\alpha) = \bigcup_{\alpha\in A}f^{-1}(V_\alpha)\).
\(f^{-1}(V_\alpha)\) is open in \(X\) for all \(\alpha \in A\). Since \(K\) is compact, \(\exists \alpha_1, \cdots, \alpha_m \in A[K\sube \bigcup_{i = 1}^m f^{-1}(V_{\alpha_i}) = f^{-1}(\bigcup_{i = 1}^m V_{\alpha_i})]\).
And hence, \(f(K) \sube \bigcup_{i = 1}^m V_{\alpha_i}\), which means \(f(K)\) is compact. □
Theorem 5.14 Given \(f:K(\sube X) \to \R\). If \(K\) is compact, then \(\max\{f(x)|x\in K\}\) and \(\min\{f(x)|x\in K\}\) exist.
Proof By Proposition 5.13, \(f(K)\) is a compact set in \(\R\). By Heine-Borel theorem, \(f(K)\) is bounded and closed.
\(\sup f(K)\) and \(\inf f(K)\) exist since \(f(K)\) is bounded.
And \(\sup f(K)\) and \(\inf f(K)\) are both accumulation points.
By Proposition 7.9, \(f(K)\) is closed, then \(\sup f(K),\inf f(K) \in f(K)\).
i.e., \(\max f(K) = \sup f(K),\min f(K) = \inf f(K)\). □
Theorem 5.15 (the intermediate value theorem) Let \(f:[a,b] \to \R\) be a continuous function. \(\forall L\in(f(a),f(b))\exists c\in [a,b][L=f(c)]\).
Proof Without loss of generality, suppose that \(f(a) < f(b)\).
Let \(a_0 = a, b_0 = b\).
Consider \(f\left(\frac{a+b}{2}\right)\). If:
- \(f\left(\frac{a+b}{2}\right) = L\), then \(\frac{a+b}{2}\) is the point we are looking for.
- \(f\left(\frac{a+b}{2}\right) < L\), let \(a_1 = \frac{a+b}{2}, b_1 = b\)
- \(f\left(\frac{a+b}{2}\right) > L\), let \(a_1 = a, b_1 = \frac{a+b}{2}\).
Repeating the operation, we may obtain \([a_n, b_n]\) which is nested intervals with \(f(a_n) < L < f(b_n) (n\in\N)\).
By theorem of nested intervals, \(\bigcap_{n=1}^\infty [a_n, b_n] = \{c\}\) for some \(c\).
We claim that \(f(c) = L\). □