Prove that the following function is $C^{\infty}$

Question

Prove that the following function: $$r:x \mapsto \begin{cases} e^{-{1\over (1-x^2)}}, & \text{if $|x|<1$} \\ 0, & \text{if $|x| \ge 1$} \end{cases}$$ is $C^{\infty}$

I found this problem on internet and i was interested to find a proof but i did't find any otheρ similar with this exercise.It would be very nice if we can have a proof for this.

Answer 1

Consider $f(x) = e^{-\frac{1}{(1 - x^2)}}$

Then note that if $f^{n}(x) = \frac{P(x)}{Q(x)}f(x)$ then $f^{n+1}(x) = \frac{P'(x) Q(x) - P(x)Q'(x)}{Q(x)^2} + (-\frac{1}{1 - x^2})'f(x) = \frac{\hat{P}(x)}{\hat{Q}(x)}f(x)$. Where $f^n$ is the n th derivative of $f$ and $P,Q, \hat{P}, \hat{Q}$ are polinomials.

Note that $Q(x) = \bigg(\frac{1}{1-x^2}\bigg)^k$ for some $k$

Now we just have to prove that $\frac{P(x)}{Q(x)} e^{-\frac{1}{(1 - x^2)}}$ is a $C^\infty$ function. On the interval $(-1,1)$ it certainly is $C^{\infty}$ outside $[-1,1]$ it is the constant fuction $0$ so it is $C^{\infty}$.

We will show that $\lim_{x \to 1} \frac{P(x)}{Q(x)} e^{-\frac{1}{(1 - x^2)}} = 0$ for general $P,Q$ and therefore the result will follow:

note that $x^2 \to 1$ is equivalent to $1 - x^2 = \delta \to 0$ ($\delta >0$ since $x \in(-1,1)$ )

$$ e^{-1/\delta} = \frac{1}{e^{1/\delta}} = \frac{1}{\sum_{l = 1}^\infty(1/\delta)^l} \leq \frac{1}{(1/\delta)^{k+ 1}} $$

$P(x) = \sum_{i = 0}^n a_i x^i \leq \sum_{i = 0}^n |a_i| = A_P$

$Q(x) = \bigg(\frac{1}{\gamma}\bigg)^k$

therefore

$$\bigg|\frac{P(1 - \delta)}{Q(1 - \delta)} e^{-\frac{1}{(1 - x^2)}} \bigg| \leq A_P \bigg(\frac{1}{\delta}\bigg)^k \frac{1}{e^{1/\delta}} \xrightarrow[\delta \to 0]{} 0$$

Hope this helps

Answer 2

It is enough to show $f:t\mapsto\left\{ \begin{array} [c]{c}% e^{-\frac{1}{1-t}}\text{ if }t<1\\ 0\text{ if }t\geq1 \end{array} \right. $ is infinitely differentiable at $1$.

Step 1: It is easy to verify that $$ \lim_{t\uparrow1}\frac{f\left( t\right) }{\left( 1-t\right) ^{k}}=0\text{ }\forall k=0,1,2,... $$

Step 2: $f^{\left( k\right) },$ the k-th derivative of $f,$ is the finite linear combination of $\frac{f\left( t\right) }{\left( 1-t\right) ^{j}% },j=0,1,2,...$ This could be verified by induction.

Answer 3

Consider the function \begin{eqnarray} \chi(x) = \begin{cases} e^{-\frac{1}{x}} \ & \text{ if}\ x> 0 \\ 0 & \text{ if}\ x \le 0 \end{cases} \end{eqnarray} then $f(x) = \chi(1-|x|^2)$, so it's enough to show that $\chi$ is $C^{\infty}$. We'll break the proof into several easy steps.

1.If $f$, $g$ are continuous functions on $\mathbb{R}$ and $f' = g$ on $\mathbb{R} \backslash\{0\}$ then $f'=g$ on $\mathbb{R}$.

Only need to look at $f'(0)$. By Lagrange mean value theorem $\frac{f(t) - f(0)}{t-0}= f'(c_t) = g(c_t)$ for some $c_t$ between $0$ and $t$, so as $t\to 0$, $c_t \to 0$, now use the continuity of $g$.

2. For $x \ne 0$ we have $$\left(e^{-\frac{1}{x}}\right)^{(n)}= R_n(x)\cdot e^{-\frac{1}{x}}$$

where $R_n(x)$ is a rational fraction in $x$. This is an easy induction.

3. Define \begin{eqnarray} \chi_n(x) = \begin{cases} \left(e^{-\frac{1}{x}}\right)^{(n)} \ & \text{ if}\ x> 0 \\ 0 & \text{ if}\ x \le 0 \end{cases} \end{eqnarray}

Then $\chi_n$ is continuous. We only need the limit at $0$ from the right, apply l'Hospital using 2.

4. We have $\chi_n' = \chi_{n+1}$. Indeed, this is true over $\mathbb{R} \backslash\{0\}$, now apply 1. using 3.

Conclude: $(\chi)^{(n)} = \chi_0^{(n)} = \chi_n$ for all $n$. Hence $\chi$ is $C^{\infty}$.