If $f(x)$ is smooth and odd, must $f(x)/x$ be smooth?

Question

Let $f:\mathbb R\to\mathbb R$ be:

1. smooth, i.e., infinitely differentiable, and
2. odd, i.e., $f(x)=-f(-x)$ for all $x$.

Let $g:\mathbb R\to\mathbb R$ be defined as

$$g(x):=\left\{ \matrix{f(x)/x, & x\neq 0 \\ \lim_{x\to 0} f(x)/x, & x=0} \right.$$

Must $g$ be smooth? In fact, is $g$ necessarily well defined at $x=0$?

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This is not a HW problem, by the way. I am have been working on a different problem and just realized I have been assuming the above fact without proof. I am aware of some theorems in complex analysis that make this true for complex-analytic functions, but I also know that real analysis isn't always so clean.

Answer 1

$g$ will always be well defined at $0$ because:

$$g(0) = \lim_{x\to 0} \frac{f(x)}{x} = \lim_{x\to 0} \frac{f(x)-0}{x-0} \equiv f'(0)$$

by the definition of the derivative. One can even prove via repeated use of quotient rule and induction that by this definition,

$$g^{(n)}(0) = \frac{f^{(n+1)}(0)}{(n+1)}$$

which is well defined because $f$ is smooth.

Answer 2

Assume that $f:\mathbb{R}\to\mathbb{R}$ is smooth and $f(0) = 0$. If we define $g$ as in OP's construction, then $g$ is well-defined and the following representation holds for all $x \in \mathbb{R}$.

$$ g(x) = \int_{0}^{1} f'(xt) \, \mathrm{d}t. $$

Then it is easy to prove that Leibniz integral rule is applicable here, showing that $g$ is indefinitely differentiable and

$$ g^{(n)}(x) = \int_{0}^{1} \frac{\partial^n}{\partial x^n} f'(xt) \, \mathrm{d}t = \int_{0}^{1} t^n f^{(n+1)}(xt) \, \mathrm{d}t. $$

In particular, we immediately find that $g^{(n)}(0) = \int_{0}^{1} t^n f^{(n+1)}(0) \, \mathrm{d}t = \frac{f^{(n+1)}(0)}{n+1}$.

Answer 3

I accepted Ninad Munshi's answer because it's correct and led me to the full proof, but I will put the full proof here for posterity:

Combining the formulas $$\left( \frac{d}{dx}\right)^k\left(\frac{1}{x}\right)=\frac{(-1)^k k!}{x^{k+1}}$$ and $$\left( \frac{d}{dx}\right)^n \big( p(x)q(x) \big)=\sum_{k=0}^n \frac{n!}{k!(n-k)!}p^{(k)}(x)q^{(n-k)}(x),$$

We have, for $x\neq 0$:

$\begin{align} \\g^{(n)}(x) &\equiv \left( \frac{d}{dx}\right)^n \left( \frac{1}{x}\cdot f(x)\right) \\ \\ \\ &= \sum_{k=0}^n \frac{n!}{k!(n-k)!}\cdot \frac{(-1)^k k!}{x^{k+1}} f^{(n-k)}(x) \\\\\\ &= \frac{1}{x^{n+1}}\sum_{k=0}^n \frac{n!}{(n-k)!}\cdot (-1)^k \cdot x^{n-k} f^{(n-k)}(x).\\ \\ \end{align}$

Now assume the formula $g^{(n)}(0)=\frac{f^{(n+1)}(0)}{n+1}$ holds for some $n$, and we hope to then show that it holds for $n+1$. By definition of the derivative, we have:

$\begin{align} \\ g^{(n+1)}(0) &= \lim_{x\to 0}\frac{g^{(n)}(x)-g^{(n)}(0)}{x-0} \\\\\\ &= \lim_{x\to 0}\frac{\left(\sum_{k=0}^n \frac{n!}{(n-k)!}\cdot (-1)^k \cdot x^{n-k} f^{(n-k)}(x) \right)-x^{n+1}\frac{f^{(n+1)}(0)}{n+1}}{x^{n+2}} \\\\ \end{align}$

The only term in the numerator without a factor of $x$ in front is the term with $f^{(0)}(x)$, which must also vanish as $x\to 0$ because $f$ is odd, so we are justified in using L'Hopital's rule. When we do, the sum telescopes, leaving only:

$\begin{align} \\ g^{(n+1)}(0) &= \lim_{x\to 0}\frac{x^n f^{(n+1)}(x)-x^n f^{(n+1)}(0)}{(n+2)x^{n+1}} \\\\\\ &= \lim_{x\to 0}\frac{f^{(n+1)}(x)-f^{(n+1)}(0)}{(n+2)x}\\\\ \end{align}$

And then one more application of L'Hopital's rule shows the formula holds for $n+1$. Ninad's post already clearly shows the formula indeed holds for $n=0$, so we are done. Interestingly, we only needed to assume that $f$ is smooth and $f(0)=0$, not that $f$ is odd.

Answer 4

Are purely real functions with purely real arguments not just a subset of the complex functions? If a theorem holds for the complex functions I see no reason it cannot hold for the real. Let $f:\mathbb{C}\rightarrow\mathbb{C}$ be a function from the complex to the complex on which your theorem $X$ applies. Because $X$ applies, certain properties of $f$ hold true. Now suppose you restricted the domain to just the reals, since that is all you care about, $f:\mathbb{R}\rightarrow\mathbb{C}$. Does theorem $X$ fall apart now when it previously held for all of $\mathbb{C}$, and if so why? Now suppose that by some happy coincidence, over this restricted domain, $\mathrm{Ran}(f)=\mathbb{R}\subset\mathbb{C}$. Thus your function now is in fact $f:\mathbb{R}\rightarrow\mathbb{R}$. Am I supposed to believe that theorem $X$ does not apply?

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I observe that, suppose, if $f(0)\ne 0$ be some finite real value, then, being smooth and continuous, $\lim f(x)\not\rightarrow 0$ also. The expression $f(0)/0$ is undefined, and in the case of the limit, $\lim f(x)/x$, it will be divergent. This would imply a discontinuity at $x=0$ in the $g(x)$ function. Would it not? Making it non-smooth.

It would seem that a counter example to your assertion would be to find a smooth, odd $f(x)$ such that $f(0)\ne 0$. But no such odd function exists that is smooth, because, intuitively, $f(+\epsilon)>0$ and $f(-\epsilon)=-f(+\epsilon)<0$ but $f(0)\ne 0$. We have, for any $f(0)=c>0$, for example, there would exist a neighborhood around $c$ whose preimage would be around 0. But for a sufficiently small neighborhood around c, $f(x)$ remains positive, despite the fact that $f(-\delta)=-f(\delta)$. This would only make sense if contradicting $f(0)\ne 0$.

Of course this isnt a proof, just an observation I ran with.

Answer 5

First, let's notice that the problem is equivalent to $g^{(n)}(0)$ being defined for every $n \geq 0$, since this shows that $g$ is differentiable to all orders, and hence has continuous derivatives of all orders.

So, we only need to prove $g^{(n)}(0)$ exists. This is not too hard if we use Taylor's theorem.

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For $x \neq 0$ and for some $n \geq 0$ $$g(x) = \frac{f(x)}{x} = \frac{f(0) + xf'(0) + \frac{1}{2}x^2f''(0) + \cdots + \frac{1}{n!}x^nf^{(n)}(0) + \frac{1}{(n+1)!}x^{n+1}f^{(n)}(0) + x^{n+1}R_n(x)}{x}$$$$ = f'(0) + \frac{1}{2}xf''(0) + \frac{1}{n!}x^{n-1}f^{(n)}(0) + \frac{1}{(n+1)!}x^{n}f^{(n)}(0) + x^{n}R_n(x)$$ where $R_n(x)$ is some smooth function with $R_n(0) = 0$. But, this is exactly the Maclaurin expansion of $g(x)$, so the derivatives of $g$ must be $$g^{(n)}(0) = \frac{1}{n+1}f^{(n+1)}(0).$$

This can be shown by induction. Certainly, $g(0) = g^{(0)}(0) = f'(0)$ (by taking $x \to 0$ in the above equation). Now, suppose the result holds true for $n$. Then, $$g^{(n)}(x) = \frac{d^n}{dx^n} \left(f'(0) + \frac{1}{2}xf''(0) + \cdots \frac{x^n}{(n+1)!}f^{(n+1)}(0)+ \frac{1}{(n+2)!}x^{n+1}f^{(n+2)}(0) + x^{n+1}R_{n+1}(x)\right)$$ $$= \frac{1}{n+1}f^{(n+1)}(0) + \frac{x}{n+2}f^{(n+2)}(0) + \frac{d^{n}}{dx^n}\left(x^{n+1}R_{n+1}(x)\right)$$ Now, $$\frac{d^n}{dx^n}\left(x^{n+1}R_{n+1}(x)\right) = (n+1)!x R_{n+1}(x) + O(x^2)$$ so, $$g^{(n+1)}(0) = \lim_{x \to 0} \frac{g^{(n)}(x) - g^{(n)}(0)}{x} = \lim_{x \to 0}\frac{1}{n+2}f^{(n+2)}(0) + R_{n+1}(x) + O(x) = f^{(n+2)}(0)$$

using the fact that $R_{n+1}(0) = 0$.