SQUARES, CUBES, AND SMOOTH FUNCTIONS

TERENCE TAO

Abstract. A little puzzle: if f : ℝ → [0, +∞) is such that f^{2} and f^{3} are both smooth, does this imply f is itself smooth?

1. Introduction

I learnt this question from Brian Conrad. It is well known that square roots of smooth functions need not be smooth. More precisely, if f : ℝ → [0, +∞) is a function such that f^{2} is smooth, this does not imply that f is smooth; indeed, the function f(x) = |x| provides an easy counterexample. The same problem occurs even if f vanishes to infinite order; consider for instance the function defined by

f(x) = [sin^{2}(\frac{1}{|x|}) + e^{−1/|x|^{2}}]^{1/2}e^{−1/|x|}

for x ≠ 0, with f(0) = 0, vanishes to infinite order at 0 and has a smooth square, but f is still not smooth at the origin (indeed, the second derivative of f is unbounded near zero).

However, suppose one assumes that f : ℝ → [0, +∞) is such that f^{2} and f^{3} are both smooth; this of course implies that all higher integer powers of f are smooth. Does this now imply that f itself is smooth? I do not know the answer to this question. If f vanishes to some finite order at a point x_{0} then it is not hard to show that f is smooth near x_{0}, and of course when f is non-zero the claim is obvious, so it is not hard to reduce matters to studying matters near a zero of infinite order. But I do not see how to prove or disprove that the derivatives of f stay bounded near such a zero. The best I can do is to show that f is locally Lipschitz, which can be established as follows. We work in the region where f is non-zero. Since f^{3} is smooth, its third derivative is locally bounded. This third derivative is

6(f′)^{3} + 18ff′f′′ + 3f^{2}f′′′.

Similarly, the third derivative of f^{2}, which is 2ff′′′ + 6f′f′′, is also locally bounded. Multiplying the latter by \frac{3}{2}f and subtracting from the former, we conclude that

6(f′)^{3} + 9ff′f′′

is locally bounded. We rewrite this as

\frac{9}{2}f′(f^{2})′′ − 3(f′)^{3}.

Since (f^{2})′′ is locally bounded, we conclude that −3(f′)^{3} + O(|f′|) is locally bounded, which easily implies that f′ itself is locally bounded.

Unfortunately it does not appear that this type of argument can be extended to show that, say, f′′ is bounded, though I cannot yet prove or disprove this. It may also be that this is too “local” of an argument; one may need to exploit the positivity to do better.

I have not seriously thought much further about this puzzle, but perhaps someone with more time on their hands might want to play with it.

2. Update (Feb 16, 2007)

The above problem was solved in the affirmative in ^{\underline{square}} [4]. (Thanks to Catalin Badea for the reference.) It is indeed possible to continue the above argument as follows.

Proposition 2.1 (Main identity). Let n ≥ 1, and suppose that m ≥ 2(n + 1). Then there is a pointwise identity of the form

(f^{(n)})^{m} = ∑_{a≥2;k≥0} P_{a,k,n,m}(f, f′, …, f^{(n)})D^{k}(f^{a})  (2.1) fnm

whenever f is smooth, where D^{k} denotes k-fold differentiation, and P_{a,k,n,m} is an explicit polynomial combination of f, f′, …, f^{(n)} (with coefficients independent of f).

Proof. By translation invariance it suffices to prove this identity at the origin. Use Taylor expansion to write f = g + h, where g is a polynomial of degree at most n, and h vanishes to order n+1 or larger at the origin. Observe that the left-hand side of (^{\underline{fnm}} 2.1) is equal to an explicit constant times

D^{nm}(g^{m})(0).

We then expand out g^{m} = (f − h)^{m} using the binomial formula as a combination of powers of f and h, with the net power adding up to m. But observe that as h vanishes to order n + 1 at the origin, any expression involving more than nm/(n + 1) powers of h will automatically vanish. Thus we only need to consider terms in which h is raised to power at most nm/(n + 1), and hence f is raised to power at least m/(n + 1). In particular, f is raised to power at least 2, leading to an identity of the form

D^{nm}(g^{m})(0) = ∑_{a≥2} c_{a,n,m}D^{nm}(f^{a}h^{m−a})(0)

for some explicit constants c_{a,n,m}. Writing h = (f − g) and using the binomial formula again, we thus obtain an identity of the form

D^{nm}(g^{m})(0) = ∑_{a≥2} c′_{a,n,m}D^{nm}(f^{a}g^{m−a})(0)

for some slightly different constants c′_{a,n,m}. Applying the Leibnitz rule, we can expand D^{nm}(f^{a}g^{m−a}) as a combination of terms of the form D^{k}(f^{a}) times some polynomial combination of derivatives of g. But the derivatives of g are in turn just f, f′, …, f^{(n)} (followed by a lot of zeroes). The claim follows. □

Now we can establish the boundedness of f^{(n)} as inductively as follows. Assume that f, f′, …, f^{(n−1)} is already known to be bounded on the region where f is non-zero. From homogeneity we see that the total degree of P_{a,k,n,m} in (^{\underline{fnm}} 2.1) can be taken to be m − a. Also since f^{a} is smooth for a ≥ 2 we see that D^{k}(f^{a}) is bounded. Thus we see that

(f^{(n)})^{m} = O(∑_{a=2}^{m} |f^{(n)}|^{m−a})

whence f^{(n)} = O(1) as desired.

References

amen [1] I. Ameniya, K. Masuda, On Joris’ theorem on differentiability of functions, Kodai Math. J. 12

(1989), 92–97.

duncan [2] J. Duncan, S. Krantz, H. Parks, Nonlinear conditions for differentiability of functions, J. Analyse

Math. 45 (1985), 46–68.

myers [3] R. Myers, An elementary proof of Joris’s theorem, Amer. Math. Monthly 112 (2005), no. 9, 829–

831.

square [4] H. Joris, Une C^{∞}-application non-immersive que possède la propriété universelle des immersions, Arch. Math. (Basel) 39 (1982), no. 3, 269-277.

Department of Mathematics, UCLA, Los Angeles CA 90095-1555, USA.

E-mail address: tao@math.ucla.edu