11.28 Continuous Function

Composition

Theorem

Let \(f: Y \to \mathbb{R}\), \(g: X \to Y\), \(g(X) \subseteq Y\). If \(f, g\) is continuous on \(Y, X\), then \(f \circ g: X \to \mathbb{R}\) is continuous on \(X\).

That is the composition of two continuous functions is also continuous

Proof

Let \(x_0 \in X\), \(y_0 = g(x_0)\).

(*)\(f\) continuous at \(y_0 \implies \forall \varepsilon > 0, \exists \delta > 0, |y - y_0| < \delta \implies |f(y) - f(y_0)| < \varepsilon.\)

\(g\) continuous at \(x_0\implies\text{Let }\varepsilon=\delta,\exists\tilde{\delta}>0,|x-x_0|<\tilde{\delta}\implies|g(x)-g(x_0)|<\delta.\)

Now since \(g(x_0)=y_0\), use (*), \(|g(x)-g(x_0)|<\delta\implies|f(g(x))-f(g(x_0))|<\varepsilon\)​\(\implies |f\circ g(x) - f\circ g(x_0)| < \varepsilon\)

Which means \(f \circ g\) is continuous at \(x_0\). Since \(x_0\) is arbitrary on \(X\), \(f \circ g\) is continuous on \(X\).

Parenthesis

\(\lim_{x \to x_0} f(x) = \lim_{h \to 0} f(x_0 + h).\)

Proof

Use composition

\(g(h) = x_0 + h, \quad \lim_{h \to 0} g(h) = x_0.\)

\(\lim_{h\to0}f(g(h))=\lim_{x\to x_0}f(x)\)

Example of continuous function

1. \(\exp(x) = \lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n = e^x = \sum_{k=0}^\infty \frac{x^k}{k!}\)

   Proposition: \(f(x) = \exp(x) = e^x\) is continuous on \(\mathbb{R}\).

   Proof

   N.T.P. \(\lim_{x\to x_0}\exp(x)=\exp(x_0)\Rightarrow\lim_{x\to x_0}e^{x}=e^{x_0}\)​

   Use parenthesis: \(\lim_{h \to 0} \exp(x_0 + h) = e^{x_0}\)​

   We know that \(e^{x_0 + h} = e^{x_0} \cdot e^h \implies \lim_{h \to 0} e^{x_0 + h} = e^{x_0} \cdot \lim_{h \to 0} e^h\), then we need to prove \(\lim_{h\to0}e^{h}=1\)

   But we don't have good tools to calculate exact the expansion of \(e^h\), thus we need to try to use this, then we need \(xg(x)\)​

   \(\exp(x)=\sum_{k=0}^{\infty}\frac{x^{k}}{k!}=1+\sum_{k=1}^{\infty}\frac{x^{k}}{k!}=1+x\sum_{k=1}^{\infty}\frac{x^{k-1}}{k!}=1+x\sum_{k=0}^{\infty}\frac{x^{k}}{\left(k+1\right)!}\).

   Then we have \(\exp(x)=1+xg(x)\)​

   Now \(\lim_{x \to x_0} \exp(x) = 1 \iff \lim_{x \to x_0} x \cdot g(x) = 0\).

   From previous class:

   If \(|g(x)| < M\) in \(x \in (x_0 - \delta_0, x_0 + \delta_0)\)

   \(\implies \lim_{x \to x_0} (x - x_0) \cdot g(x) = 0\).

   We want to use \(\lim_{x \to x_0} \exp(x)\) at \(x_0 = 0\).

   We need to show that \(g\) is bounded in an interval \((0 - \delta_0, 0 + \delta_0)\).

   \(|g(x)| = \left| \sum_{k=0}^{\infty} \frac{x^k}{(k+1)!} \right| \leq \sum_{k=0}^{\infty} \frac{|x|^k}{(k+1)!}\)

   Let \(\delta_0=1,|x|<1,x\in(-1,1)\)​

   \(\sum_{k=0}^{\infty}\frac{1}{\left(k+1\right)!}\leq\sum_{k=0}^{\infty}\frac{1}{k!}=e<3=M\).

   Now we apply \(x_0=0,\,M=3,\,\delta_0=1\) and we obtain \(\lim_{x \to 0} x \sum_{k=0}^\infty \frac{x^k}{(k+1)!} = 0\)​\(\implies \lim_{h \to 0} e^h = 1\)

2. \(\sin: \mathbb{R} \to \mathbb{R}\) is continuous

   Proof

   \(\lim_{x_0 \to 0} \sin(x_0 + h) = \sin(x_0)\)

   \(\lim_{x_0 \to 0} \sin x_0 \cos h + \cos x_0 \sin h = \lim_{x_0 \to 0} \sin x_0\)

   N.T.P. \(\lim_{h \to 0} \sin(h) = 0\) \(\lim_{h \to 0} \cos(h) = 1\)

   .... to be continued

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