Exercises

The case $x = y$ is trivial, so let $x < y$. Since $f$ is continuous on $[x,y] \subset [a,b]$ and differentiable on $(x,y) \subset (a,b)$, the Mean Value Theorem applies on $[x,y]$: for some $c \in (x,y)$, $f(y) - f(x) = f'(c)(y - x)$.

Taking absolute values, $\lvert f(y) - f(x) \rvert = \lvert f'(c) \rvert \cdot \lvert y - x \rvert \le M\lvert y - x \rvert$. The case $x > y$ is the same with the roles swapped, so the inequality holds for all $x, y \in [a,b]$.

This is the Lipschitz condition, and uniform continuity follows with $\delta = \varepsilon / M$ (any $\delta > 0$ if $M = 0$).

(a) $x_0 = 1 > 0$, and if $x_n > 0$ then $x_{n+1}$ is the average of two positive numbers, hence positive — by induction $x_n > 0$ for all $n$. (b) By AM-GM, $x_{n+1} = \dfrac{x_n + 2/x_n}{2} \ge \sqrt{x_n \cdot \dfrac{2}{x_n}} = \sqrt{2}$.

(c) $x_1 = \dfrac{1 + 2}{2} = \dfrac32$, and $\left(\dfrac32\right)^2 = \dfrac94 > 2$, so $x_1 > \sqrt{2}$ — the first step overshoots the root.

For $n \ge 1$, $x_{n+1} - x_n = \dfrac{2/x_n - x_n}{2} = \dfrac{2 - x_n^2}{2x_n}$; by (b), $x_n \ge \sqrt{2}$, so the numerator is $\le 0$. Hence $(x_n)_{n \ge 1}$ is decreasing.

(d) $(x_n)_{n \ge 1}$ is decreasing and bounded below by $\sqrt{2}$, so it converges to some $L \ge \sqrt{2}$.

Passing to the limit in the recurrence gives $L = \dfrac{L + 2/L}{2}$, i.e. $2L = L + 2/L$, so $L = 2/L$. Since $L > 0$, $L^2 = 2$, i.e. $L = \sqrt{2}$.

For $0 \le t \le 1$ and $x, y \in \mathbb{R}$, the triangle inequality and homogeneity of the absolute value give $\lvert tx + (1-t)y \rvert \le \lvert tx \rvert + \lvert (1-t)y \rvert = t\lvert x \rvert + (1-t)\lvert y \rvert$. This is exactly the definition of convexity, so $f$ is convex on $\mathbb{R}$.

A convex function is continuous on every open interval (the theorem of the convex scene), so $f$ is continuous everywhere.

But the difference quotient at $0$ is given by the formula below, which has no limit. Hence $f$ is not differentiable at $0$ — the simplest example of a function that is convex and continuous but not differentiable.

By Darboux's theorem, $f'$ has the intermediate value property. Suppose $f'$ changed sign: there exist $u, v \in [a,b]$ with $f'(u) > 0 > f'(v)$.

Applying Darboux's theorem on the interval between $u$ and $v$ would produce a point $c$ with $f'(c) = 0$, contradicting the hypothesis. Hence $f'$ cannot change sign, and since it never vanishes, it is either everywhere positive or everywhere negative on $[a,b]$.

Consider the case $f' > 0$ (the negative case differs only in sign). For $x < y$, the Mean Value Theorem gives $c \in (x,y)$ with $f(y) - f(x) = f'(c)(y - x)$; since $f'(c) > 0$, we get $f(y) > f(x)$.

So $f$ is strictly increasing. Likewise $f' < 0$ makes $f$ strictly decreasing.

For $n = 0$ the formula is $f(x) = f(a) + \int_a^x f'(t)\,dt$, the Fundamental Theorem of Calculus. As the induction hypothesis, assume the formula for $n$ holds for $C^{n+1}$ functions.

For $g \in C^{n+2}$, applying it to $g$ gives $g(x) = T_n(x) + \frac{1}{n!}\int_a^x (x-t)^n g^{(n+1)}(t)\,dt$, where $T_n$ is the degree-$n$ Taylor polynomial.

Apply integration by parts to the remainder integral: with $u = g^{(n+1)}(t)$ and $dv = (x-t)^n\,dt$, we have $du = g^{(n+2)}(t)\,dt$ and $v = -\dfrac{(x-t)^{n+1}}{n+1}$.

Since $g^{(n+1)}$ and $(x-t)^{n+1}$ are continuous, parts is legitimate and \[\int_a^x (x-t)^n g^{(n+1)}(t)\,dt = \left[-\frac{(x-t)^{n+1}}{n+1}\,g^{(n+1)}(t)\right]_a^x + \frac{1}{n+1}\int_a^x (x-t)^{n+1} g^{(n+2)}(t)\,dt.\] The boundary term is $0$ at $t = x$ and gives $\dfrac{(x-a)^{n+1}}{n+1}\,g^{(n+1)}(a)$ at $t = a$ (the minus sign flips).

Multiplying by $\frac{1}{n!}$, the remainder integral becomes $\dfrac{g^{(n+1)}(a)}{(n+1)!}(x-a)^{n+1} + \dfrac{1}{(n+1)!}\int_a^x (x-t)^{n+1} g^{(n+2)}(t)\,dt$. The first term is exactly the degree-$(n+1)$ Taylor term, and the second is $R_{n+1}(x)$.

Hence $g(x) = T_{n+1}(x) + R_{n+1}(x)$, which is the formula for $n+1$. By induction the proof is complete for all $n \ge 0$.

(a) $T$ is decreasing on $[0,1]$ with $T(0) = 1/2$ and $T(1) = 1/3$, so $T([0,1]) = [1/3, 1/2] \subset [0,1]$.

For $x, y \in [0,1]$, $\lvert T(x) - T(y) \rvert = \left\lvert \dfrac{2+y-(2+x)}{(2+x)(2+y)} \right\rvert = \dfrac{\lvert x - y \rvert}{(2+x)(2+y)}$. Since $(2+x)(2+y) \ge 2 \cdot 2 = 4$, we get $\lvert T(x) - T(y) \rvert \le \dfrac14\lvert x - y \rvert$ — a contraction with constant $1/4$.

(b) $[0,1]$ is complete, so the Banach fixed point theorem (the Banach scene) gives $T$ a unique fixed point $x^\ast \in [0,1]$. The equation $x = \dfrac{1}{2+x}$ is equivalent to $x^2 + 2x - 1 = 0$, with roots $x = -1 \pm \sqrt{2}$. The only one in $[0,1]$ is $x^\ast = \sqrt{2} - 1 \approx 0.4142$.

By the same theorem, the iteration $x_{n+1} = T(x_n)$ from any $x_0 \in [0,1]$ converges to $x^\ast$ — the error shrinks by at least a factor of $1/4$ each step.

Lemma: for any polynomial $P$, $\lim_{\lvert u \rvert \to \infty} P(u)e^{-u^2} = 0$. Indeed, for $v = u^2 \ge 0$, $e^{v} \ge \dfrac{v^m}{m!}$ (one term of the exponential's Taylor expansion), so with $2m > \deg P$ we get $\lvert P(u) \rvert e^{-u^2} \le m!\,\dfrac{\lvert P(u) \rvert}{u^{2m}} \to 0$.

We claim $f^{(n)}(x) = P_n(1/x)e^{-1/x^2}$ for $x \ne 0$, for some polynomial $P_n$, by induction. For $n = 0$, $P_0 \equiv 1$.

With $u = 1/x$, \[\frac{d}{dx}\Bigl[P(u)e^{-1/x^2}\Bigr] = P'(u)\!\left(-\frac{1}{x^2}\right)e^{-1/x^2} + P(u)e^{-1/x^2}\!\left(\frac{2}{x^3}\right) = \bigl[-u^2P'(u) + 2u^3P(u)\bigr]e^{-u^2},\] and the bracket is a polynomial in $u$, completing the induction step. Hence $f$ is infinitely differentiable at $x \ne 0$.

Now $f^{(n)}(0) = 0$ by induction. The case $n = 0$ is the definition. $f'(0) = \lim_{h \to 0} \dfrac{e^{-1/h^2}}{h} = \lim_{\lvert u \rvert \to \infty} u e^{-u^2} = 0$ (lemma, with $u = 1/h$).

Assuming $f^{(n)}(0) = 0$, \[f^{(n+1)}(0) = \lim_{h \to 0}\frac{f^{(n)}(h) - f^{(n)}(0)}{h} = \lim_{h \to 0}\frac{P_n(1/h)e^{-1/h^2}}{h} = \lim_{\lvert u \rvert \to \infty} uP_n(u)e^{-u^2} = 0\] (lemma applied to $uP_n(u)$). Thus $f^{(n)}(0) = 0$ for all $n$, and $f$ is infinitely differentiable at $0$ too.

The Maclaurin series is $T(x) = \sum_{n=0}^{\infty} \dfrac{f^{(n)}(0)}{n!}x^n \equiv 0$ — it converges (to $0$) at every $x$.

But for $x \ne 0$, $f(x) = e^{-1/x^2} > 0$. So the Taylor series converges everywhere yet matches $f$ only at $x = 0$: even a smooth function need not be recovered by its Taylor series.