# Induction proof for a function being in $C^{\infty}(\mathbb{R})$

Let $$f(t) = \begin{cases} 0 & \text{: } x \le 0\\ e^{-\frac{1}{t^2}}& \text{: } x > 0 \end{cases}$$ Prove that $$f(t)$$ is in $$C^{\infty}(\mathbb{R})$$.

I know that for a function to be in $$C^{\infty}(\mathbb{R})$$ it has to be infinitely times differentiable and each derivative must be continuous. I'm a bit stuck on how to prove this. I'm assuming the answer will need to be proved inductively. Ie:

Suppose the $$f^{(n)}(t)$$ is continuous. I need to then prove that $$f^{n+1}(t)$$ exists and is continuous. I'm not quite sure how to do this, I'm new to proofs by induction. I can also image proving continuity at $$t=0$$ may be a challenge.

• Typos in your function: you either use $t$ or $x$, not both.
– user9464
Feb 5 '21 at 23:13
• It may help to notice after a handful of computations that these derivatives for $t > 0$ (which clearly exist by any number of theorems) have a common form, namely, rational function of $t$ times $e^{-1/t^2}$, and maybe trying to prove a general result about that instead of working out explicit formulas for all of the coefficients in the specific rational functions at issue. Feb 5 '21 at 23:17

Hint.

A standard example of a smooth non-analytic function is $$g(x)=\left\{\begin{array}{ll}e^{-\frac{1}{x}} & \text { if } x>0 \\ 0 & \text { if } x \leq 0\end{array}\right.$$

A detailed proof for smoothness is given in this Wikipedia article:

https://en.wikipedia.org/wiki/Non-analytic_smooth_function#The_function_is_smooth

You can adapt the techniques for your example. A key step is to note that for any positive integer $$m$$, $$\lim_{x\downarrow 0}\frac{e^{-\frac1{x}}}{x^{m}}=0$$

Alternatively, observe that $$f(x)=g(h(x))$$ where $$h(x)=x^2$$ for $$x\ge 0$$ and $$h(x)=0$$ for $$x<0$$. You can show that $$h$$ is smooth. The result then follows from the chain rule.

A simple induction argument shows that for $$x >0$$ $$f^{n}(x)$$ can be written as $$p_n(\frac 1 x) e^{-1/x^{2}}$$ for some sequence of polynomials $$(p_n)$$ But $$\lim_{y \to \infty} \frac {p_n(y)} {e^{y^{2}}} =0$$ by L'Hopital's Rule. Put $$y=\frac 1 x$$ in this.

Find all derivatives of $$f(x)$$ everywhere on $$\mathbb{R}$$ except for $$t = 0$$. Then prove that all of them in fact are continuous at zero.