Given function $f\colon \mathbb{R}\to \mathbb{R}$ be defined by $$ f(x)=\begin{cases} e^{-\frac{1}{x^2}} & \text{if $x > 0$;}\\ 0 & \text{if $x \leq 0$.} \end{cases} $$

a) Show that $g(x)=f(x-a)f(b-x)$ is a smooth function, positive on $(a,b)$ for $a<b$ and zero else where then

$$ h(x)=\frac{\displaystyle\int_{-\infty}^{x} g(t)\,dt}{\displaystyle\int_{-\infty}^{\infty} g(t)\,dt} $$

is smooth function satisfying $h(x)=0$ for $x<a$, $h(x)=1$ for $x>b$ and $0<h(x)<1$ for $x \in (a,b)$.

b) Construct a smooth function on $\mathbb{R}^k$ that equals 1 on the ball of radius $a$ and zero outside the ball of radius $b$ and is strictly between $0$ and $1$ at intermediate points $(0<a<b)$

Here is what I got so far

First I need to show that $f(x)$ is smooth.

For $x < 0$, $f(x)$ is differentiable. For $x>0$, same thing happen. For $x=0$

$\displaystyle\lim _{h\to0^-} \frac{e^{-\frac{1}{h^2}}}{h}=0$ also by L'Hospital rule

$\displaystyle\lim _{h\to0^+} \frac{e^{-\frac{1}{h^2}}}{h}=0$ by L'Hospital rule

So for $x=0$, $f(x)$ also differentiable. Now we need to check if $f'$ is continuous.

For $x>0$ and $x<0$, f(x) is continuous, by calculus rule. At $x=0$ we have

$f'(x)= -\frac {2}{x^3} e^{-\frac{1}{x^2}}$

so by L'Hospital rule

$\displaystyle \lim _{x\to0^+} -\frac {2}{x^3} e^{-\frac{1}{x^2}}=0=\lim _{x\to0^-} -\frac {2}{x^3} e^{-\frac{1}{x^2}}$

so $f'$ is continuous at $x=0$. Hence $f(x)$ is smooth. Since $f(x)$ is smooth $g(x)$ is smooth.

Now I'm stuck

  • 1
    $\begingroup$ Seems correct to me. Your function is in fact not just smooth, it is $C^\infty$ meaning that all its $n$'th derivatives exist and are continuous. $\endgroup$
    – Winther
    Sep 19, 2014 at 21:41
  • 4
    $\begingroup$ @Winther Possibly by "smooth" they mean $C^\infty$? $\endgroup$ Sep 19, 2014 at 21:43
  • 1
    $\begingroup$ @JeppeStigNielsen Oh yeah that the standard definiton. Thanks for clearing that up. Sometimes smooth it used about $C^1$ or $C^2$ (espesially in physics). But then OP has a bit more work to do: you must show that all derivatives exist and are continuous. $\endgroup$
    – Winther
    Sep 19, 2014 at 21:45
  • $\begingroup$ Hint: Show, that there exist polynomials $p_n(x)$ of degree $2n-2$ with $f^{(n)}(x) = p_n(x)x^{-3n}e^{-1/x^2}$ this should give your limits. $\endgroup$
    – andre
    Sep 20, 2014 at 0:07
  • $\begingroup$ For the polynomials $p_n(x)$ you may consult oeis.org/A215216 $\endgroup$
    – andre
    Sep 20, 2014 at 0:25

3 Answers 3


I will try to not only give you a solution but also motivate it a bit:

a) Smoothness of $f$: To show that $f$ is smooth, we have to show that all derivatives $f^{(n)}$ exist. You have already found out that $f'(x) = \frac{-2}{x^3} e^{-1/x^2}$. For $x > 0$, we then get $$f''(x) = \frac6{x^4} e^{-1/x^2} + \frac{-2}{x^3} \cdot \frac{-2}{x^3} e^{-1/x^2} = \frac{6x^2 + 4}{x^6} e^{-1/x^2}.$$ So we have, for $x > 0$, \begin{align*} f(x) &= r_0(x) \cdot e^{-1/x^2} \quad \text{with $r_0(x) = 1$},\\ f'(x) &= r_1(x) \cdot e^{-1/x^2} \quad \text{with $r_1(x) = \frac{-2}{x^3}$},\\ f''(x) &= r_2(x) \cdot e^{-1/x^2} \quad \text{with $r_2(x) = \frac{6x^2 + 4}{x^6}$}. \end{align*} The $r_i$ are rational functions (quotients of polynomials) and it is easy to see that the derivative of a rational function is again a rational function whose denominator has no additional zeros.

In the above equations there clearly is a pattern. So let us conjecture the following: For all $n \ge 0$ and all $x > 0$, $f^{(n)}(x)$ exists and has the form $f{(n)}(x) = r_n(x) \cdot e^{-1/x^2}$ where $r_n$ is a rational function whose denominator is some power of $x$.

Lets prove that by induction on $n$: The base case $n = 0$ is clear. Now assume that for some $n$ the assertion is correct. Then we get for all $x > 0$: \begin{align*} f^{(n+1)}(x) = (f^{(n)})'(x) = \left(r_n(x) \cdot e^{-1/x^2}\right)' = r_n'(x) \cdot e^{-1/x^2} + r_n \cdot \frac{-2}{x^3} e^{-1/x^2} = \left(r_n'(x) - \frac{2 r_n(x)}{x^3}\right) e^{-1/x^2}. \end{align*} Define $r_{n+1}(x)$ to be the left factor in this term. Then it is not too hard to check that $r_{n+1}$ is a rational function whose denominator is a power of $x$. In the following, let us write $r_n(x) = p_n(x) / x^{k_n}$ for some polynomial $p_n$.

So far we have shown that $f$ is smooth for every $x > 0$. Also, it is trivial that $f$ is smooth for $x < 0$ and that all derivatives are zero there. Thus it remains to be shown that $f$ is also smooth at $0$. So let us conjecture that $f^{(n)}(0)$ exists and is equal to $0$. This is true for $n = 0$. Now suppose it is true for some $n$. Then $$f^{(n+1)}(0) = \lim_{h \to 0} \frac{f^{(n)}(h)}{h} = \lim_{h \to 0} \frac{r_n(h)}{h \cdot e^{1/x^2}} = \lim_{h \to 0} \frac{p_n(h)}{h^{1+k_n} \cdot e^{1/x^2}} = 0.$$ The last step is l'Hospital's rule (applied multiple times).

Hence $f$ is smooth.

The properties of $g$ are clear. The properties of $h$ are also not too hard to see. Not that the infinite integral $\int_{\infty}^{\infty} g(x) dx$ is actually finite, since $g$ is zero outside of $(a, b)$.

b) Lets first consider the case $k = 1$. Take $F = 1 - h$ on [a, b]. Then $F$ is a smooth function with $F(x) = 1$ for $x = a$, $1 > F(x) > 0$ for $x \in (a, b)$ and $F(x) = 0$ for $x = b$. Thus $F$ has the desired properties for $x \in [a, b]$. We extend $F$ by saying $F(x) = 1$ for $|x| < a$, $F(x) = 0$ for $|x| > b$ and $F(x) = F(|x|)$ for $x \in [-b, -a]$.

Check that this $F$ is smooth. Then certainly $F$ is the function we are looking for.

Now, for a general $k$, we can try a similar thing. On $\Bbb R^k$ we have the norm $\|x\|$ which is smooth for $x \ne 0$. Using this norm and the function $F$, can you find a solution for b)?


The only part requiring some work is to prove that the function $$f(x):=\cases{\exp(-1/x)\quad&$(x>0)$\cr 0&$(x\leq0)$\cr}$$ is smooth. I claim that $$f^{(n)}(x)=\cases{p_{2n}(1/x)\exp(-1/x)\quad&$(x>0)$\cr 0&$(x\leq0)$\cr}\qquad(n\geq0)\ ,\tag{1}$$ where $p_{2n}$ denotes a certain polynomial of degree $2n$.

Proof. Obviously $(1)$ is true for $x<0$ and all $n\geq0$, and an easy induction shows that $(1)$ is true for $x>0$ and all $n\geq0$ as well. It remains to check what happens at $x=0$. We have $f^{(0)}(0)=0$ by definition; therefore assume $f^{(n)}(0)=0$ for some $n\geq0$. Then $$\lim_{x\to0-}{f^{(n)}(x)-f^{(n)}(0)\over x}=0$$ is trivial, and $$\lim_{x\to0+}{f^{(n)}(x)-f^{(n)}(0)\over x}=\lim_{x\to0+}{1\over x}p_{2n}(1/x)\exp(-1/x)=\lim_{y\to\infty}y\> p_{2n}(y)e^{-y}=0$$ as well. It follows that $f^{(n+1)}(0)=0$.$\qquad\qquad\square$

Part a) of your problem follows now from inspection, and for part b) I propose $$q({\bf x}):=1-h(\|{\bf x}\|)\ .$$


Hint for a)

First one does not need the square in the definition of $f$ to prove that $f$ is a smooth function.

Have you tried some of my advice in the comments?

Let $s(x) = e^{-1/x}$. First task would be to show that

\begin{equation} s^{(n)}(x)_{x=1/y} = \sum\limits_{k=1}^{n}\binom{n}{k-1}\frac{(n-1)!}{(n-k)!}(-y)^{2n-k+1}s(1/y) \end{equation}

For the individual monomials in $y$ of the $s^{(n)}$ term, one gets $\lim_{y\to\infty} \frac{y^r}{e^y} = 0$, which proves the smoothness.

Hint for b) Have you tried something with the euclidean distance?

  • $\begingroup$ I'm not sure I understand your suggestion. Like I said, I'm not very good at topology, and I'm self studying this book, so there will be many thing that got me confused without a detail guidance. $\endgroup$ Sep 25, 2014 at 21:37
  • $\begingroup$ But do you know how to prove something using complete induction? Base case? Induction step? And you surely know how to differentiate a polynomial and the exponential function? I'm not going to post a detailed solution. $\endgroup$
    – andre
    Sep 25, 2014 at 22:25

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