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12.5 Theorems of Continuity

If f: X \to \mathbb{R} is continuous, and [a,b] \subset X, then the image of [a,b] under f is al...

Lemma

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If \(f:X\rightarrow\mathbb{R}\) is continuous at \(x_0\in X\), then \(\exists\delta_0,M_0\) such that \(\forall |x-x_0|<\delta_0,|f(x)|<M_0\)

Informally: If \(f\) is continuous at \(x_0\), then close enough to \(x_0\), \(f\) is bounded

Proof

Since \(f\) is continuous at \(x_0\), then for \(\varepsilon=1,\exists\delta_0\) such that \(|x-x_0|<\delta_0\) such that \(|f(x)-f(x_0)|<1\), then \(|f(x)|<1+|f(x_0)|\)

Just take \(M_0=1+|f(x_0)|\), then we are done

Example

It's false if \(f\) is not continuous, let \(f(x)=\begin{cases}\frac{1}{x},x\neq0\\0,x=0\end{cases}\), always useful to give a counter example?

Then \(f\) is not continuous since it is not continuous at \(x_0=0\) ( \(\lim_{x\to0}f\left(x\right)\neq0=f\left(0\right)\) )

But this function is not bounded close enough to \(x_0=0\)

Theorem(Bolzano-Weierstrass)

Let \(f:[a,b]\rightarrow \R\) be continuous. Then \(f\) is bounded on \([a,b]\), that is \(\exists M>0\) such that \(|f(x)|<M,\forall x\in[a,b]\)

Proof

Let \(A=\{x\in[a,b]:f\text{ is bounded on }[a,x]\}\), since \(a\in A\) which is \(f\) is bounded on \([a,a]=\{a\}\) by \(|f(a)|+1\)

Use lemma above: \(\exists \delta_a\) such that \(f\) is bounded on \([a,a+\delta_a)\)

image​

Then \(f\) is bounded on \([a,a+\frac{\delta_a}{2}]\), then \(a+\frac{\delta_a}{2}\in A\)

Thus \(A\) is bounded above: \(b\) or \(b+1\) is a bounded for \(A\)

By sup axiom, let \(\alpha=\sup(A)\)

We want to prove \(b\in A\), we are going to prove \(\alpha=b\) and \(\alpha\in A\)

  1. We know \(a+\frac{\delta_a}{2}\leq\alpha\leq b\) since \(b\) is an upper bound and \(a+\frac{\delta_a}{2}\in A\)

    image​

    Suppose \(\alpha<b\), then apply the lemma to \(x_0=\alpha\)​

    Then \(\exists\delta_{\alpha}>0:f\text{ is bounded on }(\alpha-\delta_\alpha,\alpha+\delta_\alpha)\), then \(f\) is bounded on \([\alpha-\frac{\delta_{\alpha}}{2},\alpha+\frac{\delta_{\alpha}}{2}]\)​

    But since \(\alpha-\frac{\delta_\alpha}{2}<\alpha\) and \(\alpha\) is \(\sup(A)\), then \(x_1\in A:\alpha-\frac{\delta_{\alpha}}{2}<x_1<\alpha\Rightarrow f\) is bounded on \([a,x_1]\)​

    Now \(f\) is bounded on \([a,x_1]\) and \(f\) is bounded on \([\alpha-\frac{\delta_{\alpha}}{2},\alpha+\frac{\delta_{\alpha}}{2}]\), then \(f\) is bounded on \([a,\alpha+\frac{\delta_\alpha}{2}]\), which is a contradiction to \(\alpha\) is a sup

  2. Suppose \(\alpha\notin A\) and we know \(\alpha=b\)\, then \(f\) is bounded in \([\alpha-\frac{\delta_\alpha}{2},b]\) by lemma

    image​

    Since \(\alpha-\frac{\delta_{\alpha}}{2}<\alpha=\sup\left(A\right)\), then \(\exists x_1\in A\) such that \(\alpha-\frac{\delta_{\alpha}}{2}<x_1<\alpha=b\)​

    Then \(f\) is bounded on \([a,x_1]\), but \(f\) is also bounded on \([\alpha-\frac{\delta_\alpha}{2},b]\), then \(f\) is bounded on \([a,b]\)​

    Then \(b\in A\)

MAX Min Theorem

Let \(f:[a,b]\rightarrow \R\) be continuous. Then \(\exists x_1,x_2\in [a,b]\) such that \(\forall x\in[a,b],f(x_1)\leq f\left(x\right)\leq f\left(x_2\right)\). That is \(f\) has \(\max\) and \(\min\) value on \([a,b]\)

image​

Proof

Let \(A=range(f)=\{y:y=f(x)\text{ for some }x\in[a,b]\}\)

We know \(A\neq \empty\) since \(f(a),f(b)\in A\)

Also \(A\) is bounded since the \(M\) of B-W theorem is an upper bound of \(A\) and \(-M\) is a lower bound of \(A\)

Thus let \(\alpha=\inf(A)\) and \(\beta=\sup(A)\), we want to see that \(\alpha\) is \(\min\) and \(\beta=\max\)

That is we want to see \(\exists x_1\in[a,b]\) such that \(f(x_1)=\alpha\) and \(\exists x_2\in[a,b]\) such that \(f(x_2)=\beta\)

image​

Suppose \(x_2\) doesn't exist, then \(\forall x\in [a,b]:f(x)\neq \beta\)

\(g(x)=\frac{1}{\beta-f(x)}>0\)

  1. \(g\) is continuous because \(1\) and \(\beta-f(x)\) is continuous and \(\beta-f(x)\neq 0\)​

  2. Goal: Prove \(g\) is not bounded, that is \(\forall M>0,\exists x\in[a,b]:f(x)>M\)

    image​

    Let \(M>0\), then \(\beta-\frac{1}{M}<\beta\). Since \(\beta=\sup(A)\), then \(\exists y\in A\) such that \(\beta-\frac{1}{M}<y<\beta\)

    Since \(y\in A,\exists x\in[a,b]\) such that \(f(x)=y\), then \(\beta-\frac{1}{M}<f(x)\Rightarrow\frac{\beta M-1}{M}<f\left(x\right)\Rightarrow\beta M-1<Mf\left(x\right)\\\Rightarrow M\left(\beta-f\left(x\right)\right)<1\Rightarrow M<\frac{1}{\beta-f\left(x\right)}=g\left(x\right)\)

    Which contradicts to \(g\) is continuous, then \(\exists x_2\in[a,b]\) such that \(f(x_2)=\beta\)

Similarly, \(\exists x_1\in[a,b]\) such that \(f(x_1)=\alpha\)

exercise

Corollary

If \(f:[a,b]\rightarrow \R\) is continuous, then \(im(f)=[c,d]\)

Proof

Take \(c=\alpha=f(x_1),d=\beta=f(x_2)\) and use intermedia value theorem to show that \(im(f)=[c,d]\)

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