Proof of Property of Continuous Functions
3.4: Properties of Continuous Functions
- Page ID
- 49109
Recall from Definition 2.6.3 that a subset \(A\) of \(\mathbb{R}\) is compact if and only if every sequence \(\left\{a_{n}\right\}\) in \(A\) has a subsequence \(\left\{a_{n_{k}}\right\}\) that converges to a point \(a \in A\).
Let \(D\) be a nonempty compact subset of \(\mathbb{R}\) and let \(f: D \rightarrow \mathbb{R}\) be a continuous function. Then \(f{D}\) is a compact subset of \(\mathbb{R}\). In particular, \(f(D)\) is closed and bounded.
- Proof
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Take any sequence \(\left\{y_{n}\right\}\) in \(f(D)\). Then for each \(n\), there exists \(a_{n} \in D\) such that \(y_{n}=f\left(a_{n}\right)\). Since \(D\) is compact, there exists a subsequence \(\left\{a_{n_{k}}\right\}\) of \(\left\{a_{n}\right\}\) and a point \(a \in D\) such that
\[\lim _{k \rightarrow \infty} a_{n_{k}}=a \in D .\]
It now follows from Theorem 3.3.3 that
\[\lim _{k \rightarrow \infty} y_{n_{k}}=\lim _{k \rightarrow \infty} f\left(a_{n_{k}}\right)=f(a) \in f(D) .\]
Therefore, \(f(D)\) is compact.
The final conclusion follows from Theorem 2.6.5 \(\square\)
We say that the function \(f: D \rightarrow \mathbb{R}\) has an absolute minimum at \(\bar{x} \in D\) if
\[f(x) \geq f(\bar{x}) \text { for every } x \in D.\]
Similarly, we say that \(f\) has an absolute maximum at \(\bar{x}\) if
\[f(x) \leq f(\bar{x}) \text { for every } x \in D.\]
Figure \(3.2\): Absolute maximum and absolute minimum of \(f\) on \([a, b]\).
Suppose \(f: D \rightarrow \mathbb{R}\) is continuous and \(D\) is a compact set. Then \(f\) has an absolute minimum and an asolute maximum on \(D\).
- Proof
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Since \(D\) is compact, \(A=f(D)\) is closed and bounded (see Theorem 2.6.5). Let
\[m=\inf A=\inf _{x \in D} f(x).\]
In particular, \(m \in \mathbb{R}\). For every \(n \in \mathbb{N}\), there exists \(a_{n} \in A\) such that
\[m \leq a_{n}<m+1 / n.\]
For each \(n\), since \(a_{n} \in A=f(D)\), there exists \(x_{n} \in D\) such that \(a_{n}=f\left(x_{n}\right)\) and, hence,
\[m \leq f\left(x_{n}\right)<m+1 / n.\]
By the compactness of \(D\), there exists an element \(\bar{x} \in D\) and a subsequence \(\left\{x_{n_{k}}\right\}\) that converges to \(\bar{x} \in D\) as \(k \rightarrow \infty\). Because
\[m \leq f\left(x_{n_{k}}\right)<m+\frac{1}{n_{k}} \text { for every } k\]
by the squeeze theorem (Theorem 2.1.6) we conclude \(\lim _{k \rightarrow \infty} f\left(x_{n_{k}}\right)=m\). On the other hand, by continuity we have \(\lim _{k \rightarrow \infty} f\left(x_{n_{k}}\right)=f(\bar{x})\). We conclude that \(f(\bar{x})=m \leq f(x)\) for every \(x \in D\). Thus, \(f\) has an absolute minimum at \(\bar{x}\). The proof is similar for the case of absolute maximum. \(\square\)
The proof of Theorem 3.4.2 can be shortened by applying Theorem 2.6.4. However, we have provided a direct proof instead.
If \(f:[a, b] \rightarrow \mathbb{R}\) is continuous, then it has an absolute minimum and an absolute maximum on \([a, b]\).
- Proof
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Corollary 3.4.4 is sometimes referred to as the Extreme Value Theorem. It follows immediately from Theorem 3.4.2, and the fact that the interval \([a, b]\) is compact (see Example 2.6.4).
The following result is a basic property of continuous functions that is used in a variety of situations.
Let \(f: D \rightarrow \mathbb{R}\) be continuous at \(c \in D\). Suppose \(f(c) > 0\). Then there exists \(\delta > 0\) such that
\[f(x)>0 \text { for every } x \in B(c ; \delta) \cap D.\]
- Proof
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Let \(\varepsilon=f(c)>0\). By the continuity of \(f\) at \(c\), ther exists \(\delta > 0\) such that \(x \in D\) and \(|x-c|<\delta\), then
\(|f(x)-f(c)|<\varepsilon\).
This implies, in particular, that \(f(x)>f(c)-\varepsilon=0\) for every \(x \in B(c ; \delta) \cap D\). The proof is now complete. \(\square\)
An analogous result holds if \(f(c)<0\).
Let \(f:[a, b] \rightarrow \mathbb{R}\) be a continuous function. Suppose \(f(a) \cdot f(b)<0\) (this means either \(f(a)<0<f(b) \text { or } f(a)>0>f(b)\)). Then there exists \(c \in (a,b)\) such that \(f(c)=0\).
- Proof
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We prove only the case \(f(a)<0<f(b)\) (the case \(f(a)>0>f(b)\) is completely analogous). Define
\[A=\{x \in[a, b]: f(x) \leq 0\} .\]
This set is nonempty since \(a \in A\). This set is also bounded since \(A \subset[a, b]\). Therefore, \(c=\sup A\) exists and \(a \leq c \leq b\). We are going to prove that \(f(c)=0\) by showing that \(f(c)<0\) and \(f(c)>0\) lead to contradictions.
Suppose \(f(c)<0\). Then there exists \(\delta > 0\) such that
\[f(x)<0 \text { for all } x \in B(c ; \delta) \cap[a, b] .\]
Because \(c<b\) (since \(f(b)>0\)), we can find \(s \in(c, b)\) such that \(f(s) < 0\) (indeed \(s=\min \{c+ \delta / 2,(c+b) / 2\}\) will do). This is a contradiction because \(s \in A\) and \(s>c\).
Suppose \(f(c)>0\). Then there exists \(\delta > 0\) such that
\[f(x)>0 \text { for all } x \in B(c ; \delta) \cap[a, b] .\]
Since \(a<c\) (because \(f(a)<0\)), there exists \(t \in(a, c)\) such that \(f(x)>0\) for all \(x \in(t, c)\) (in fact, \(t=\max \{c-\delta / 2,(a+c) / 2\}\) will do). On the other hand, since \(t<c=\sup A\), there exists \(t^{\prime} \in A\) with \(t<t^{\prime} \leq c\). But then \(t<t^{\prime}\) and \(f\left(t^{\prime}\right) \leq 0\). This is a contradiction. We conclude that \(f(c)=0\). \(\square\)
Let \(f:[a, b] \rightarrow \mathbb{R}\) be a continuous function. Suppose \(f(a)<\gamma<f(b)\). Then there exists a number \(c \in(a, b)\) such that \(f(c)=\gamma\).
The same conclusion follows if \(f(a)>\gamma>f(b)\).
Figure \(3.3\): Illustration of the Intermediate Value Theorem.
- Proof
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Define
\[\varphi(x)=f(x)-\gamma, x \in[a, b] .\]
Then \(\varphi\) is continuous on \([a,b]\). Moreover,
\[\varphi(a) \varphi(b)=[f(a)-\gamma][f(b)-\gamma]<0 .\]
By Theorem 3.4.7, there exists \(c \in (a,b)\) such that \(\varphi(c)=0\). This is equivalent to \(f(c)=\gamma\). The proof is now complete. \(\square\)
Let \(f:[a, b] \rightarrow \mathbb{R}\) be a continuous function. Let
\[m=\min \{f(x): x \in[a, b]\} \text { and } M=\max \{f(x): x \in[a, b]\} .\]
Then for every \(\gamma \in[m, M]\), there exists \(c \in [a,b]\) such that \(f(c)=\gamma\).
- Proof
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We will use the Intermediate Value Theorem to prove that the equation \(e^{x}=-x\) has at least one real solution. We will assume known that the exponential function is continuous on \(\mathbb{R}\) and that \(e^{x}<1\) for \(x<0\).
Solution
First define the function \(f: \mathbb{R} \rightarrow \mathbb{R}\) by \(f(x)=e^{x}+x\). Notice that the given equation has a solution \(x\) if and only if \(f(x)=0\). Now, the function \(f\) is continuous (as the sum of continuous functions). Moreover, note that \(f(-1)=e^{-1}+(-1)<1-1=0\) and \(f(0)=1>0\). We can now apply the Intermediate Value Theorem to the function \(f\) on the interval \([-1,0]\) with \(\gamma=0\) to conclude that there is \(c \in [-1,0]\) such that \(f(c)=0\). The point \(c\) is the desired solution to the original equation.
We show now that, given \(n \in \mathbb{N}\), every positive real number has a positive \(n\)-th root.
Solution
Let \(n \in \mathbb{N}\) and let \(a \in \mathbb{R}\) with \(a>0\). First observe that \((1+a)^{n} \geq 1+n a>a\) (see Exercise 1.3.7). Now consider the function \(f:[0, \infty) \rightarrow \mathbb{R}\) given by \(f(x)=x^{n}\). Since \(f(0)=0\) and \(f(1+a)>a\), it follows from the Intermediate Value Theorem that there is \(x \in(0,1+a)\) such that \(f(x)=a\). That is, \(x^{n}=a\), as desired. (We show later in Example 4.3.1 that such an \(x\) is unique.)
We present below a second proof of Theorem 3.4.8 that does not depend on Theorem 3.4.7, but, instead, relies on the Nested Intervals Theorem (Theorem 2.3.3).
We construct a sequence of nested intervals as follows. Set \(a_{1}=a\), \(b_{1}=b\), and let \(I_{1}=[a, b]\). Let \(c_{1}=(a+b) / 2\). If \(f\left(c_{1}\right)=\gamma\), we are done. Otherwise, either
\[\begin{array}{l}
f\left(c_{1}\right)>\gamma \quad \text { or } \\
f\left(c_{1}\right)<\gamma
\end{array} .\]
In the first case, set \(a_{2}=a_{1}\) and \(b_{1}=c_{1}\). In the second case, set \(a_{2}=c_{1}\) and \(b_{2}=b_{1}\). Now set \(I_{2}=\left[a_{2}, b_{2}\right]\). Note that in either case,
\[f\left(a_{2}\right)<\gamma<f\left(b_{2}\right) .\]
Set \(c_{2}=\left(a_{2}+b_{2}\right) / 2\). If \(f\left(c_{2}\right)=\gamma\), again we are done. Otherwise, either
\[\begin{array}{l}
f\left(c_{2}\right)>\gamma \quad \text { or } \\
f\left(c_{2}\right)<\gamma
\end{array} .\]
In the first case, set \(a_{3}=a_{2}\) and \(b_{3}=c_{2}\). In the second case, set \(a_{3}=c_{2}\) and \(b_{3}=b_{2}\). Now set \(I_{3}=\left[a_{3}, b_{3}\right]\). Note that in either case,
\[f\left(a_{3}\right)<\gamma<f\left(b_{3}\right) .\]
Proceeding in this way, either we find some \(c_{n_{0}}\) such that \(f\left(c_{n_{0}}\right)=\gamma\) and, hence, the proof is complete, or we construct a sequence of closed bounded intervals \(\left\{I_{n}\right\}\) with \(I_{n}=\left[a_{n}, b_{n}\right]\) such that for all \(n\),
- \(I_{n} \supset I_{n+1}\),
- \(b_{n}-a_{n}=(b-a) / 2^{n-1}\), and
- \(f\left(a_{n}\right)<\gamma<f\left(b_{n}\right)\).
In this case, we proceed as follows. Condition (ii) implies that \(\lim _{n \rightarrow \infty}\left(b_{n}-a_{n}\right)=0\). By the Nested Intervals Theorem (Theorem 2.3.3, part(b)), there exists \(c \in[a, b]\) such that \(\bigcap_{n=1}^{\infty} I_{n}=\{c\}\). Moreover, as we see from the proof of that theorem, \(a_{n} \rightarrow c\) and \(b_{n} \rightarrow c\) as \(n \rightarrow \infty\).
By the continuity of \(f\), we get
\[\begin{array}{l}
\lim _{n \rightarrow \infty} f\left(a_{n}\right)=f(c) \quad \text { and } \\
\lim _{n \rightarrow \infty} f\left(b_{n}\right)=f(c)
\end{array} .\]
Since \(f\left(a_{n}\right)<\gamma<f\left(b_{n}\right)\) for all \(n\), condition (iii) above and Theorem 2.1.5 give
\[\begin{array}{l}
f(c) \leq \gamma \quad \text { and } \\
f(c) \geq \gamma
\end{array} .\]
It follows that \(f(c)=\gamma\). Note that, since \(f(a)<\gamma<f(b)\), then \(c \in (a,b)\). The proof is now complete. \(\square\)
Now we are going to discus the continuity of the inverse function. For a function \(f: D \rightarrow E\), where \(E\) is a subset of \(\mathbb{R}\), we can define the new function \(f: D \rightarrow \mathbb{R}\) by the same function notation. The function \(f: D \rightarrow E\) is said to be continuous at a point \(\bar{x} \in D\) if the corresponding function \(f: D \rightarrow \mathbb{R}\) is continuous at \(\bar{x}\).
Let \(f:[a, b] \rightarrow \mathbb{R}\) be strictly increasing and continuous on \([a,b]\). Let \(c=f(a)\) and \(d=f(b)\). Then \(f\) is one-to-one, \(f([a, b])=[c, d]\), and the inverse function \(f^{-1}\) defined on \([c,d]\) by
\[f^{-1}(f(x))=x \text { where } x \in[a, b] ,\]
is a continuous function from \([c,d]\) onto \([a,b]\).
- Proof
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The first two assertions follow from the monotonicity of \(f\) and the Intermediate Value Theorem (see also Corollary 3.4.9). We will prove that \(f^{-1}\) is continuous on \([c,d]\). Fix any \(\bar{y} \in [c,d]\) and fix any sequence \(\left\{y_{k}\right\}\) in \([c,d]\) that converges to \(\bar{y}\). Let \(\bar{x} \in[a, b]\) and \(x_{k} \in[a, b]\) be such that
\[f(\bar{x})=\bar{y} \text { and } f\left(x_{k}\right)=y_{k} \text { for every } k .\]
Then \(f^{-1}(\bar{y})=\bar{x}\) and \(f^{-1}\left(y_{k}\right)=x_{k}\) for every \(k\). Suppose by contradiction that \(\left\{x_{k}\right\}\) does not converge to \(\bar{x}\). Then there exists \(\varepsilon_{0}>0\) and a subsequence \(\left\{x_{k_{\ell}}\right\}\) of \(\left\{x_{k}\right\}\) such that
\[\left|x_{k_{\ell}}-\bar{x}\right| \geq \varepsilon_{0} \text { for every } \ell .\]
Since the sequence \(\left\{x_{k_{\ell}}\right\}\) is bounded, it has a further subsequence that converges to \(x_{0} \in[a, b]\). To simplify the notation, we will again call the new subsequence \(\left\{x_{k_{\ell}}\right\}\). Taking limits in (3.7), we get
\[\left|x_{0}-\bar{x}\right| \geq \varepsilon_{0}>0 .\]
On the other hand, by the continuity of \(f\), \(\left\{f\left(x_{k_{\ell}}\right)\right\}\) converges to \(f\left(x_{0}\right)\). Since \(f\left(x_{k_{\ell}}\right)=y_{k_{\ell}} \rightarrow \bar{y}\) as \(\ell \rightarrow \infty\), it follows that \(f\left(x_{0}\right)=\bar{y}=f(\bar{x})\). This implies \(x_{0}=\bar{x}\), which contradicts (3.8). \(\square\)
A similar result holds if the domain of \(f\) is the open interval \((a,b)\) with some additional considerations. If \(f:(a, b) \rightarrow \mathbb{R}\) is increasing and bounded, followin the argument in Theorem 3.2.4 we can show that both \(\lim _{x \rightarrow a^{+}} f(x)=c\) and \(\lim _{x \rightarrow b^{-}} f(x)=d\) exist in \(\mathbb{R}\) (see Exercise 3.2.10). Using the Intermediate Value Theorem we obtain that \(f((a, b))=(c, d)\). We can now proceed as in the previous theorem to show that \(f\) has a continuous inverse from \((c,d)\) to \((a,b)\).
If \(:(a, b) \rightarrow \mathbb{R}\) is increasing, continuous, bounded below, but not bounded above, then \(\lim _{x \rightarrow a^{+}} f(x)= c \in \mathbb{R}\), but \(\lim _{x \rightarrow b^{-}} f(x)=\infty\) (again see Exercise 3.2.10). In this case we can show using the Intermediate Value Theorem that \(f((a, b))=(c, \infty)\) and we can proceed as above to prove that \(f\) has a continuous inverse from \((c, \infty)\) to \((a,b)\).
The other possibilities lead to similar results.
A similar theorem can be proved for strictly decreasing functions.
Exercise \(\PageIndex{1}\)
Let \(f: D \rightarrow \mathbb{R}\) be continuous at \(c \in D\) and let \(\gamma \in \mathbb{R}\). Suppose \(f(c)>\gamma\). Prove that there exists \(\delta>0\) such that
\[f(x)>\gamma \text { for every } x \in B(c ; \delta) \cap D .\]
- Answer
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Exercise \(\PageIndex{2}\)
Let \(f,g\) be continuous functions on \([a,b]\). Suppose \(f(a)<g(a)\) and \(f(b)>g(b)\). Prove that there exists \(x_{0} \in(a, b)\) such that \(f\left(x_{0}\right)=g\left(x_{0}\right)\).
Exercise \(\PageIndex{3}\)
Prove that the equation \(\cos x=x\) has at least one solution in \(\mathbb{R}\). (Assume known that the function \(\cos x\) is continuous.
Exercise \(\PageIndex{4}\)
Prove that the equation \(x^{2}-2=\cos (x+1)\) has at least two real solutions. (Assume known that the function \(\cos x\) is continuous.)
Exercise \(\PageIndex{5}\)
Let \(f:[a, b] \rightarrow[a, b]\) be a continuous function.
- Prove that the equation \(f(x)=x\) has a solution on \([a,b]\).
- Suppose further that
\[|f(x)-f(y)|<|x-y| \text { for all } x, y \in[a, b], x \neq y .\]
Prove that the equation \(f(x)=x\) has a unique solution on \([a,b]\)
Exercise \(\PageIndex{6}\)
Let \(f\) be a continuous function on \([a,b]\) and \(x_{1}, x_{2}, \ldots, x_{n} \in[a, b]\). Prove that there exists \(c \in [a,b]\) with
\[f(c)=\frac{f\left(x_{1}\right)+f\left(x_{2}\right)+\cdots f\left(x_{n}\right)}{n}.\
Exercise \(\PageIndex{7}\)
Suppose \(f\) is a continuous function on \(\mathbb{R}\) such that \(|f(x)| < |x| \text { for all } x \neq 0\).
- Prove that \(f(0)=0\).
- Given two positive numbers \(a\) and \(b\) with \(a < b\), prove that there exists \(\ell \in[0,1)\) such that \[|f(x)| \leq \ell|x| \text { for all } x \in[a, b].\]
Exercise \(\PageIndex{8}\)
Let \(f, g:[0,1] \rightarrow[0,1]\) be continuous functions such that
\[f(g(x))=g(f(x)) \text { for all } x \in[0,1] . \nonumber\]
Suppose further that \(f\) is monotone. Prove that there exists \(x_{0} \in[0,1]\) such that
\[f\left(x_{0}\right)=g\left(x_{0}\right)=x_{0} . \nonumber\]
Source: https://math.libretexts.org/Bookshelves/Analysis/Introduction_to_Mathematical_Analysis_I_(Lafferriere_Lafferriere_and_Nguyen)/03%3A_Limits_and_Continuity/3.04%3A_Properties_of_Continuous_Functions
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