Calculus

Theorem 1 (⚠ Intermediate Value Theorem) Suppose that \(f : [a,b] \to \mathbb R\) is continuous. Then, \(f\) is bounded on \([a,b]\) and for all \(y \in [\min_{[a,b]} f, \max_{[a,b]} f ]\) there exists \(x\in[a,b]\) such that \(f(x) = y\).

Theorem 2 (⚠ Mean Value Theorem) Suppose that \(f \colon [a,b] \to \mathbb R\) is continuous on \([a,b]\) and differentiable on \((a,b)\). Then, there exists \(c\in[a,b]\) for which

\[\begin{align*} f(b) - f(a) = f'(c) (b-a) \end{align*}\]

Proof: Consider \(g(x) := f(x) - \frac{f(b)-f(a)}{b-a} x\), a function for which \(g(a) = g(b)\). You can show that there exists \(c\in [a,b]\) for which \(g'(c) = 0\) (this is known as Rolle’s theorem). This follows from the intermediate value theorem - either the max or min is achieved at \(a\) and \(b\) (and the function is constant), or there is an extreme point in \((a,b)\). In the latter case, you can show this point is stationary: i.e. \(g'\) vanishes here.

Cauchy’s Mean Value Theorem (need this for the proof of Taylor remainder theorem): Suppose that \(f, g \colon [a,b] \to \mathbb R\) are continuous on \([a,b]\) and differentiable on \((a,b)\). Then, there exists \(c\in[a,b]\) for which

\[\begin{align*} \big( f(b) - f(a) \big) g'(c) = \big(g(b) - g(a) \big) f'(c) \end{align*}\]

Proof: \(h(x) = \big( g(b) - g(a) \big) f(x) - \big( f(b) - f(a) \big) g(x)\) is such that \(h(a) = h(b)\) and so there exists \(c \in [a,b]\) such that \(h'(c) = 0\).

Theorem 3 (⚠ Taylor Remainder Theorem) Suppose that \(f\colon [a,b] \to \mathbb R\) is a function with \(n\) times continuously differentiable. Then, there exists \(c\in[a,b]\) for which

\[\begin{align*} f(x) = f(a) + f'(a) (x - a) + \dots + \frac{f^{(n-1)}(a)}{(n-1)!} (x - a)^{n-1} + \frac{f^{(n)}(c)}{n!} (x-a)^{n} \end{align*}\]

Proof: Define \(F(x) = f(x) - \sum_{k=0}^{n-1} \frac{f^{(k)}(a)}{k!} (x - a)^k\) and \(G(x) = (x-a)^n\) and note that \(F(a) = F'(a) = \cdots = F^{(n-1)}(a) = 0\) and \(G(a) = \cdots = G^{(n-1)}(a) = 0\) and \(G^{(n)}(x) = n!\). Applying Cauchy’s Mean Value Theorem \(n\) times, we can conclude there exists \(c_1,\dots,c_n\) between \(a\) and \(x\) such that

\[\begin{align*} \frac{F(x)}{G(x)} &= \frac{F(x) - F(a)}{G(x) - G(a)} = \frac{F'(c_1)}{G'(c_1)} \nonumber\\ &= \frac{F'(c_1) - F'(a)}{G'(c_1) - G'(a)} = \frac{F''(c_2)}{G''(c_2)} \nonumber\\ &= \cdots = \frac{F^{(n)}(c_n)}{G^{(n)}(c_n)} % = \frac {f^{(n)}(c_n)} { n! } \end{align*}\]

Theorem 4 (⚠ Change of Sign Theorem) Suppose that \(f\colon [a,b] \to \mathbb R\) is a continuous function with \(f(a) f(b) \leq 0\). Then, \(f\) has a root (or zero) \(\xi \in [a,b]\) (i.e. \(f(\xi) = 0\)).

Proof. If \(f(a) f(b) = 0\) then \(a\) or \(b\) is a root of \(f\). Otherwise, \(0 \in (\min_{[a,b]}f, \max_{[a,b]}f)\) and so the Intermediate Value Theorem applies.

Theorem 5 (Brouwer’s Fixed Point Theorem) Suppose that \(g \colon [a,b] \to [a,b]\) is continuous. Then, there exists a fixed point \(\xi \in [a,b]\).

Proof. We can apply the change of sign theorem to \(f(x) := g(x) - x\).

Theorem 6 (Contraction Mapping Theorem (Banach fixed point theorem)) Suppose \(g:[a,b] \to[a,b]\) is a contraction. Then, there exists a unique fixed point \(\xi = g(\xi) \in [a,b]\). Moreover, the iteration \(x_{n+1} = g(x_n)\) converges at least linearly to \(\xi\) for all \(x_1 \in [a,b]\).

Notice that \(|x_{n+1} - \xi| = |g(x_n) - g(\xi)| \leq L |x_n - \xi|\) for all \(n \geq 1\). Therefore, we have \(|x_{n+1} - \xi| \leq L^n |x_1 - \xi|\) which goes to zero as \(n\to\infty\) (as \(|L|<1\)). Finally, we note that by the Mean Value Theorem there exist \(\eta_n\) between \(x_n\) and \(\xi\) such that

\[\begin{align} \frac {|x_{n+1} - \xi|} {|x_n - \xi|} % = \frac {|g(x_{n}) - g(\xi)|} {|x_n - \xi|} = |g'(\eta_n)|. \end{align}\]

This quantity converges to \(|g'(\xi)|\) as \(n \to \infty\).