12.2 Bolzano, Intermedia value theorem

Bolzano's Theorem

Let \(f: [a, b] \to \mathbb{R}\) be a continuous function s.t. \(f(a) < 0 < f(b)\). Then there is \(c \in (a, b)\) such that \(f(c) = 0\).

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Proof

Consider \(A=\{x\in\left\lbrack a,b\right\rbrack:f(t)<0,\forall t\in[a,x]\}\subseteq[a,b]\)​

1. \(a \in A\). In fact, since \(f(a)<0,\ \exists\ \delta_{a}:f(t)<0\ ,\forall t\in\left\lbrack a,a+\delta_{a}\right)\)

   In particular, \(a+\frac{\delta_{a}}{2}\in A\) because \(f<0\text{ in }[a,a+\frac{\delta_{a}}{2}]\)

2. \(b\) is UB of \(A\). Moreover, since \(f(b)>0,\exists\ \delta_{b}\ s.t.\ f(t)>0\ ,\forall t\in(b-\delta_{b},b]\)

   In particular, \(b-\frac{\delta_{b}}{2}\) is UB of \(A\).

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Lemma: If \(f: X \to \mathbb{R}\) is continuous in \(x_0\) and \(f(x_0) > 0\), \(f(x_0) \neq 0\), \(f(x_0) < 0\), then \(\exists \delta > 0\) such that \(\forall t \in (x_0 - \delta, x_0 + \delta) \cap X\), we have \(f(t) > 0\), \(f(t) \neq 0\), \(f(t) < 0\).

Proof

Since \(f\) is continuous, then \(\forall\varepsilon,\exists\delta>0,\left|x-x_0\right|<\delta:\left|f\left(x\right)-f\left(x_0\right)\right|<\varepsilon\)

If \(f(x_0)>0\), then take \(\varepsilon=\frac{f\left(x_0\right)}{2}\)...

Actually we can always take \(\varepsilon=\frac{\left|f\left(x_0\right)\right|}{2}\)​

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By the SUP axiom, we know that \(A\) has a supremum.

Let \(c = \sup(A)\) (with \(c>a+\frac{\delta_{a}}{2}\) and \(c<b-\frac{\delta_{b}}{2}\Rightarrow c\in\left(a,b\right)\)). Let us prove that \(f(c) = 0\).

We know that there are three possibilities: \(f(c) = 0\), \(f(c) < 0\), \(f(c) > 0\).

Let us see why \(f(c) > 0\) is impossible.

If \(f(c) > 0\), \(\implies \exists \delta_c > 0\) such that \(f > 0\) in \((c - \delta_c, c + \delta_c)\).

However, since \(c-\delta_{c}<c=\sup(A)\)​\(\implies c - \delta_c\) is not an upper bound of \(A\).

Thus, \(\exists x_0 \in A\) such that \(c - \delta_c < x_0 < c\), then \(f < 0\) in \([a, x_0]\), \(f > 0\) in \((c - \delta_c, x_0]\),
which is impossible for \(f(c) > 0\).

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If \(f(c) < 0\), \(\implies \exists \delta_c > 0\) such that \(f < 0\) in \((c - \delta_c, c + \delta_c)\).

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Thus \(f < 0\) in \(\left[c - \frac{\delta_c}{2}, c + \frac{\delta_c}{2}\right]\).

Since \(c - \frac{\delta_c}{2} < c = \sup(A)\)​\(\implies c - \frac{\delta_c}{2}\) is not an upper bound of \(A\).

\(\implies \exists x_0 \in A : c - \frac{\delta_c}{2} < x_0 < c\). Thus we have \(f < 0\) in \([a, x_0]\) and \(f < 0\) in \(\left[c - \frac{\delta_c}{2}, c + \frac{\delta_c}{2}\right]\).

\(\implies f<0\text{ in }[a,c+\frac{\delta_{c}}{2}]\text{ and }c+\frac{\delta_{c}}{2}\in A\). This is impossible since \(c = \sup(A)\) and \(c + \frac{\delta_c}{2} > c\).

Therefore, the only possibility left is \(f(c) = 0\).

Corollary

If \(f : [a, b] \to \mathbb{R}\) is continuous and \(f(a) > 0>f(b)\)​\(\implies \exists c \in (a, b)\) such that \(f(c) = 0\).

Proof

Take \(g(x) = -f(x)\)(since continuous)\(\implies g(a) = -f(a) < 0\), \(g(b) = -f(b) > 0\). Apply Bolzano's theorem \(\implies \exists c \in (a, b)\) such that \(g(c) = 0 \implies f(c) = 0\).

Intermediate Value Theorem

If \(f : [a, b] \to \mathbb{R}\) is continuous, then\(\forall y_0\) in between \(f(a)\) and \(f(b)\), \(\exists x_0 \in (a, b)\) such that \(f(x_0) = y_0\).

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Proof: Assume \(f(a) < f(b)\) and \(f(a) < y_0 < f(b)\).

Define \(g(x) = f(x) - y_0\). Continuous

We have \(g(a) = f(a) - y_0 < 0\), \(g(b) = f(b) - y_0 > 0\).

Apply Bolzano theorem to \(g\): \(\implies \exists c \in (a,b)\) s.t. \(g(c) = 0 \implies f(c) - y_0 = 0\)​\(\implies\) for \(x_0 = c\) we have \(f(x_0) = y_0\)

Corollary

If \(f: X \to \mathbb{R}\) is continuous, and \([a,b] \subset X\), then the image of \([a,b]\) under \(f\) is also an interval (Image of closed intervals are also closed intervals).

Proof: next lecture

Corollary

If \(f: [a,b] \to [c,d]\) is continuous and bijective, then \(f\) is strictly monotonic. (not necessary bijective, injective is enough)

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Proof: \(f\) bijective \(\implies f(a) \neq f(b)\).

2 Possibilities:

1. \(f(a) < f(b)\) \(\implies f\) is strictly monotonic increasing.

   Suppose NOT \(\implies \exists x_1 < x_2\) s.t. \(f(x_1) \geq f(x_2)\), since bijective, then \(f(x_1)>f(x_2)\)​

   Subcase (i) :
   ​\(f(x_1) > f(x_2) > f(a)\). Let \(y_0 = f(x_2)\).

   By IVThm \(\exists x_0 \in (a, x_1)\): \(f(x_0) = y_0 = f(x_2)\).

   \(\implies\) Contradiction because \(f\) is bijective.

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   Subcase (ii) :
   ​\(f(x_1) > f(a) > f(x_2)\)

   Apply IVThm to \(y_0 = f(a)\)​\(\implies\exists x_0\in(x_1,x_2)\) s.t. \(f(x_0) = f(a)\)

   \(a < x_1 < x_0 < x_2\) \(\implies\) Contradiction because \(f\) is bijective.

   Subcases (i) and (ii) cover the possibilities for \(f(x_1) > f(a)\).

   Now suppose \(f(x_1) < f(a)\)​\(\implies f(x_1) < f(b)\)

   Subcase (iii) :
   ​\(f(x_2) < f(x_1) < f(b)\)

   Apply IVThm for \(y_0 = f(x_1)\)​\(\implies\exists x_0\in(x_2,b)\) s.t. \(f(x_0) = f(x_1)\)

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   \(x_1<x_2<x_0<b\) \(\implies\) Contradiction since \(f\) is bijective.

   Therefore: Impossible \(f(x_1) > f(x_2)\) \(\implies f(x_1) < f(x_2)\)​\(\implies f\) is M.I. (strictly monotonic increasing).

2. If \(f(a) > f(b)\) \(\Rightarrow\) similarly you get \(f~MD\).

In (a) \(f\) is \(MI\), in (b) \(f\) is \(MD\) and in any case is strict since \(f\) is bijective (we don't have \(f(x_1) = f(x_2) \Rightarrow x_1 \neq x_2\)).

Theorem

If \(f:[a,b]\rightarrow[c,d]\) is continuous and bijective, then \(f^{-1}\) is continuous.

Proof

We know that \(f\) is monotonic. Say that \(f\) is \(MI\).

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We want to show that \(f^{-1}:[c,d]\rightarrow[a,b]\) is continuous.

We want to show that \(\forall y_0 \in [c, d]\), \(f^{-1}\) is continuous at \(y_0\)

We want to show that \(\forall\epsilon>0,\ \exists\ \delta>0\ s.t.\ |y-y_0|<\delta\Rightarrow|f^{-1}\left(y\right)-f^{-1}(y_0)|<\epsilon\).

We know that \(f\) is continuous.

Let \(x_0 = f^{-1}(y_0)\)

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Since monotonic increasing, let \(y_1=f(x_0-\varepsilon)<f\left(x_0\right)=y_0\) and \(y_2=f(x_0+\varepsilon)>f\left(x_0\right)=y_0\)​

Let \(\delta=\min\left(|y_1-y_0|,|y_2-y_0|\right)>0\), where \(y_0 - y_1 > 0\), \(y_2 - y_0 > 0\).

This \(\varepsilon\) works: If \(|y-y_0|<\delta\), then \(y_0-\delta<y<y_0+\delta\).

Then \(y_0-\delta\leq y_1<y<y_2\leq y_0+\delta\).

I claim that \(f^{-1}\) is also MI, then \(f^{-1}\left(y_0-\delta\right)\leq f^{-1}\left(y_1\right)<f^{-1}\left(y\right)<f^{-1}\left(y_2\right)\leq f^{-1}\left(y_0+\delta\right)\)​

Then \(x_0-\varepsilon<f^{-1}(y)<x_0+\varepsilon\Rightarrow|f^{-1}(y)-f^{-1}(y_0)|<\varepsilon\)​

To finish, we show that \(f^{-1}\) is MI (monotonically increasing):

If \(y_1 < y_2\), then \(\begin{cases}f^{-1}(y_1)<f^{-1}(y_2)\text{ only possibility }\\ f^{-1}\left(y_1\right)\geq f^{-1}\left(y_2\right)\Rightarrow f\left(f^{-1}\left(y_1\right)\right)\geq f\left(f^{-1}\left(y_2\right)\right)\Rightarrow y_1\geq y_2\end{cases}\)

Example

Example: Prove that \(\sqrt[3]{x}\) is cont. from \(\mathbb{R} \to \mathbb{R}\).

Let \(g(x) = x^3\). Is \(g: \mathbb{R} \to \mathbb{R}\) bijective?

Yes: \(g\) is MI: If \(x_1 < x_2\), then \(x_1^3 < x_2^3\) (from first classes).

We proved that \(\lim_{x \to \infty} x^3 = \infty\) (1) \(\lim_{x \to -\infty} x^3 = -\infty\) (2)

Let us see that \(\forall y_0 \in \mathbb{R}\) \(\exists x_0 \in \mathbb{R}\) s.t. \(x^3 = y_0\)

(1) \(\Rightarrow \exists \alpha > 0\) s.t. \(x>\alpha\Rightarrow x^3>y_0\Rightarrow\left(\alpha+1\right)^3>y_0\)

(2) \(\Rightarrow \exists \beta < 0\) s.t. \(x<\beta\Rightarrow x^3<y_0\Rightarrow\left(\beta-1\right)^3<y_0\)

\(\Rightarrow (\beta-1)^3 < y_0 < (\alpha+1)^3\)

We have \(g(\beta - 1) < y_0 < g(\alpha + 1)\).

By the IVT (Intermediate Value Theorem): \(\exists \, x_0 \in (\beta - 1, \alpha + 1) \, \text{s.t.} \, g(x_0) = y_0 \Rightarrow x_0^3 = y_0\) \(\Rightarrow g: \mathbb{R} \to \mathbb{R}\) is bijective

\(\sqrt[3]{}\overset{\text{def}}{=}g^{-1}\) exists, which is cont. because \(g\) is cont.

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