$\S 1$ - Smooth Functions on a Euclidean Space

1.1. A function that is $C^2$ but not $C^3$

Let $g: \mathbb{R} \to \mathbb{R}$ be the function in Example 1.2(iii). Show that the function $h(x) = \int_0^x g(t) \, dt$ is $C^2$ but not $C^3$ at $x = 0$.

Solution

First, recall that $g: \mathbb{R} \to \mathbb{R}$ is defined as
$$g(x) = \frac{3}{4}x^{4/3}$$

So
$$g'(x) = x^{1/3}$$

And
$$g''(x) = \begin{cases} \frac{1}{3}x^{-2/3} & \text{for } x \ne 0, \\ \text{undefined} & \text{for } x = 0. \end{cases}$$
By the fundamental theorem of calculus $h'(x) = g(x)$, so this shows that $h$ is $C^2$ but not $C^3$ at $x = 0$ (as $h''' = g''$ ). Note that $h$ is a polynomial and therefore continuous.

1.2.* A $C^\infty$ function very flat at 0

Let $f(x)$ be the function on $\mathbb{R}$ defined in Example 1.3.

(a)

Show by induction that for $x > 0$ and $k \geq 0$, the $k$-th derivative $f^{(k)}(x)$ is of the form $p_{2k}(1/x)e^{-1/x}$ for some polynomial $p_{2k}(y)$ of degree $2k$ in $y$.

Solution

First recall that $f(x)$ is defined on $\mathbb{R}$ by
$$f(x) = \begin{cases} e^{-1/x} & \text{for } x > 0, \\ 0 & \text{for } x \leq 0 \end{cases}$$
The base case of the induction is for $k = 0$. By definition when $x > 0$ we have $f(x) = e^{-1/x} = 1 \cdot e^{-1/x}$, proving the hypothesis via the polynomial $p_0(y) = 1$ which is of degree $2k = 0$.

Now assume that $f^{(k)}(x) = p_{2k}(1/x)e^{-1/x}$. We need to show that $f^{(k+1)}(x) = p_{2k+2}(1/x)e^{-1/x}$ for some polynomial $p_{2k+2}$ of degree $2k + 2$. Proceed by differentiating
$$\begin{aligned} f^{(k+1)}(x) &= \frac{d}{dx} \left( p_{2k}\left(\frac{1}{x}\right) e^{-1/x} \right) \\ &= \frac{d}{dx} \left( p_{2k}\left(\frac{1}{x}\right) \right) e^{-1/x} + p_{2k}\left(\frac{1}{x}\right) \frac{d}{dx} e^{-1/x} \\ &= p_{2k}'\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right) e^{-1/x} + p_{2k}\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right) e^{-1/x} \\ &= -\frac{1}{x^2} \left( p_{2k}'\left(\frac{1}{x}\right) + p_{2k}\left(\frac{1}{x}\right) \right) e^{-1/x} \end{aligned}$$
Since $p_{2k}$ is of degree $2k$, $p'_{2k}$ is of degree $2k - 1$, so $\left( p_{2k}'\left(\frac{1}{x}\right) + p_{2k}\left(\frac{1}{x}\right) \right)$ is of degree $2k$, and $p_{2k+2} = -\frac{1}{x^2} \left( p_{2k}'\left(\frac{1}{x}\right) + p_{2k}\left(\frac{1}{x}\right) \right)$ is of degree $2k + 2$.

(b)

Prove that $f$ is $C^\infty$ on $\mathbb{R}$ and that $f^{(k)}(0) = 0$ for all $k \geq 0$.

Solution

To prove that $f$ is $C^{\infty}$ on $\mathbb{R}$ it is sufficient to show that it is smooth at $x = 0$, as we handled that case $x > 0$ in part (a) and the case $x < 0$ is trivial as we have a whole epsilon ball where $f = 0$. Take $k \geq 0$, we need to show that the limit
$$\lim_{h \to 0} \frac{f^{(k)}(h) - f^{(k)}(0)}{h}$$
exists, and for the second part of the question we need to show that it equals to zero. We proceed by induction. For the base case $k = 0$ we have that
$$\lim_{h \to 0} \frac{f^{(k)}(h) - f^{(k)}(0)}{h} = \lim_{h \to 0} \frac{f(h) - f(0)}{h}$$
This splits into two cases: approaching from the left and from the right. From the left we always have $h < 0$ so
$$\lim_{h \to 0^-} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^-} \frac{0 - 0}{h} = 0$$
From the right we always have $h > 0$ so
$$\lim_{h \to 0^+} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^+} \frac{e^{-1/h} - 0}{h}$$
Let $y = 1/h$ so as $h \to 0^+$ we have $y \to \infty$
$$\lim_{h \to 0^+} \frac{e^{-1/h} - 0}{h} = \lim_{y \to \infty} ye^{-y} = \lim_{y \to \infty} \frac{y}{e^y} = 0$$
The last equality follows from L'Hospital's Rule. Now it's time for the induction. Assume that
$$\lim_{h \to 0} \frac{f^{(k)}(h) - f^{(k)}(0)}{h} = 0$$
We need to show that (note that the following limit is not even assumed to exist)
$$\lim_{h \to 0} \frac{f^{(k+1)}(h) - f^{(k+1)}(0)}{h} = 0$$
From the assumption it is clear that $f^{(k+1)}(0) = 0$ so all that remains to show is that
$$\lim_{h \to 0} \frac{f^{(k+1)}(h)}{h} = 0$$
If $h$ approaches from the left than this is obvious as the left side of the function is identically zero (since the left side of $f$ is zero so are all of its derivatives). So without loss of generality approach from the right, and use part (a)
$$\lim_{h \to 0^+} \frac{f^{(k+1)}(h)}{h} = \lim_{h \to 0^+}\frac{p_{2k+2}(1/h)e^{-1/h}}{h} = \lim_{h \to 0^+} \frac{p_{2k+3}(1/h)}{e^{1/h}}$$
Once again substitute $y = 1/h$
$$\lim_{h \to 0^+} \frac{f^{(k+1)}(h)}{h} = \lim_{y \to \infty} \frac{p_{2k+3}(y)}{e^{y}} = 0$$
The last equality following from repeated applications of L'Hospital's Rule.

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