I have a question $f(x)$ is even and can be infinitely differentiable, how about $f(\sqrt{x})$ in [0,$\infty$)? can we say that the $f(\sqrt{x})$ also can be infinitely differentiable in $[0,\infty)$.

My thoughts: let $g(x)=f(\sqrt{x})$, and then I have proved $g(x)$ can be differentiable and continuous in $[0,\infty)$, and I supposed this conclusion still holds for $n=k$($k$ is an integer),which means $g(x)$ can be $k$-times differentiable and continous in $[0,\infty)$, so I want to use mathematical induction to prove it. But I am stuck here, since I have no idea how to prove g can be $(k+1)$ times differentiable using the assumption that $g$ can be $k$-times differentiable and continuous. Can someone tell me whether it is true or not?

  • $\begingroup$ this peoblem is from my professor $\endgroup$
    – python3
    Feb 1, 2015 at 3:38
  • 1
    $\begingroup$ I think it is true, but I am struggling for it $\endgroup$
    – python3
    Feb 1, 2015 at 3:42
  • 1
    $\begingroup$ Have you tried using the chain rule? $\endgroup$
    – Math1000
    Feb 1, 2015 at 3:51
  • 2
    $\begingroup$ The proof is trivial for analytic functions. I wonder if some argument by continuity could reach the smooth case (we'd have to choose some unusual metric/topology to make such an argument work though) $\endgroup$ Feb 1, 2015 at 3:55
  • 1
    $\begingroup$ $limit_{x->0} {{f(\sqrt{x})-f(0)}\over x}$ $\endgroup$
    – python3
    Feb 1, 2015 at 4:08

1 Answer 1


Suppose $f$ is even and infinitely differentiable, and let $g(x) = f(\sqrt{x})$. I claim that for each positive integer $k$, $g^{(k)}$ is defined on $[0, \infty)$, and moreover that $g^{(k)}(x^2)$ is an infinitely differentiable even function on $\Bbb{R}$.

Suppose that the claim holds for $k=1$, and letsuppose that $g^{(n)}(x)$ is defined for $x \ge 0$ and that $g^{(n)}(x^2)$ is even and infinitely differentiable. Then, we may define $h(x) = g^{(n)}(\sqrt{x})$, and from the base case we find that $h$ is differentiable and $h'(x^2)$ is even and infinitely differentiable. But $h(x) = g^{(n)}(x)$, so $g^{(n)}$ satisfies the required conditions. It follows that $g$ is infinitely differentiable, so long as we can prove that $g'(x)$ exists for $0 \le x$ and $g'(x^2)$ is infinitely differentiable. Let's try to prove that.

We have that for $x > 0$,

$$g'(x) = \frac{f'(\sqrt{x})}{2\sqrt{x}}$$

by the chain rule, and for $x=0$,

$$g'(0) = \lim_{h\to0} \frac{g(h) - g(0)}{h} = \lim_{h\to0} \frac{f(\sqrt{h}) - g(0)}{h}= \lim_{h\to0} \frac{f(0) + f'(0)\sqrt{h} + \frac{1}{2}f''(0)h + o(h) - f(0)}{h}$$

by Taylor's Theorem. Now, since $f$ is even, $f'(0) = 0$, so

$$g'(0) = \frac{1}{2}f''(0)$$

Define $h(x) = g'(x^2)$. Then,

$$h(x) = \begin{cases} \frac{f'(x)}{2x} & x \not= 0 \\ \frac{f''(0)}{2} & x = 0 \end{cases}$$

Since $f$ is even, we have at once that $f'(x)$ is odd, so $\frac{f'(x)}{2x}$ is even and $h$ is even.

I claim that $h$ is infinitely differentiable, with

$$h^{(k)}(x) = \begin{cases}\frac{1}{2x^{k}}\sum_{n=0}^k (-1)^{n+k} \frac{(k-1)!}{n!} x^n f^{(n+1)}(x) & x \not= 0\\ \frac{f^{(k+2)}(0)}{2(k+1)} & x=0\end{cases}$$

In the base case, the identity holds for $k=0$. Now, assuming the identity holds for $k$, we have that for $x\not=0$

$$h^{(k+1)}(x) = \frac{-k}{2x^{k+1}}\sum_{n=0}^k (-1)^{n+k} \frac{(k-1)!}{n!} x^n f^{(n+1)}(x) + \frac{x}{2x^{k+1}}\sum_{n=1}^k (-1)^{n+k} \frac{(k-1)!}{(n-1)!} x^{n-1} f^{(n+1)}(x)$$ $$+ \frac{x}{2x^{k+1}}\sum_{n=0}^k (-1)^{n+k} \frac{(k-1)!}{n!} x^n f^{(n+2)}(x)$$ $$= \frac{1}{2x^{k+1}}\sum_{n=0}^{k+1} (-1)^{n+k+1} \frac{k!}{n!} x^n f^{(n+1)}(x)$$ Since the terms $n=1,2,\ldots k$ in the second sum cancel with the terms $n=0,1,\ldots,k-1$ in the third sum. And for $x=0$, we Taylor series expand each of the $f^{(m)}$ terms in $h^{(k)}$ to find that for $x \not=0$,

$$\begin{align*} h^{(k)}(x) &= \frac{1}{2x^{k+1}} \sum_{n=0}^{k} (-1)^{n+k} \frac{k!}{n!} x^n f^{(n+1)}(x)\\ &= \frac{(-1)^k}{2x^{k+1}} \sum_{n=0}^k (-1)^n \frac{k!}{n!} x^n \sum_{m=n}{^k+2} \frac{f^{(m+1)}(0)x^{m-n}}{(m-n)!} + o(x^{k+2})\\ &= \frac{(-1)^k}{2x^{k+1}}\left(\sum{m=0}^{k+2} f^{(m+1)}(0)x^m \frac{k!}{m!} \sum_{n=0}^m \frac{(-1)^n m!}{n!(m-n)!}\right) + \frac{f^{(k+2)}(0)}{2(k+1)} \\&\qquad+ \frac{xf^{(k+3)}(0)}{2}\left[\frac{1}{k+1} - \frac{1}{(k+1)(k+2)}\right] + o(x)\\ &= \frac{f^{(k+2)}(0)}{2(k+1)} + \frac{xf^{(k+3)}(0)}{2(k+2)} + o(x) \end{align*}$$

Thus, $$h^{(k+1)}(0) = \lim_{x\to0}\frac{h^{(k)}(x) - h^{(k)}(0)}{x} = \frac{f^{(k+3)}(0)}{2(k+2)}$$

This establishes the desired result.

  • $\begingroup$ I don't think that your formula for $h'$ is correct. Also, it would probably be easier to begin the induction for $h^{(k)}$ by $k=0$, but your formula does not hold for this case. $\endgroup$
    – PhoemueX
    Feb 1, 2015 at 9:36
  • $\begingroup$ @PhoemueX I've edited my answer. $\endgroup$
    – user88319
    Feb 1, 2015 at 23:28
  • $\begingroup$ looks like I have understood what you mean, thank you for your idea! $\endgroup$
    – python3
    Feb 2, 2015 at 6:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.