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Here is a prototype of the problem I have in mind: Let $P:\mathbb{R}^2\rightarrow\mathbb{R}$ be a strictly convex, nonnegative polynomial such that $P(0,0)=0$. Let $\alpha\geq 0$, and consider the following version of its distribution function (in the sense of, say, harmonic analysis)

$$\lambda_P(\alpha)=|\{(x,y)\in\mathbb{R}^2: P(x,y)<\alpha\}|;$$

here by $|\cdot|$ I mean Lebesgue measure on the plane.

Question. Is $\lambda=\lambda_P$ a smooth ($C^\infty$) function of $\alpha$, for $\alpha>0$?

Thank you.

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Don't you mean $|P(x,y)|<\alpha$? Otherwise $\lambda_P(\alpha)$ is going to be infinite for $\alpha>1$ and $P$ nonconstant. –  Alex Becker Jul 5 '12 at 0:53
    
@Alex: Corrected, thanks. I guess one could equivalently ask about the function $$\lambda_P(\alpha)=|\{(x,y)\in K: |P(x,y)|>\alpha\}|,$$ where $K$ is some compact and convex subset of the plane... –  user17240 Jul 5 '12 at 1:04
1  
Since you assume strict convexity, why take the modulus? It would be more natural to consider the set where $P<\alpha$. Also, are you interested in $C^1$ or higher degrees of smoothness? –  user31373 Jul 5 '12 at 1:27
    
@Alex: What polynomial are you taking? $P(x,y)=x^2+y^2$? Then $\lambda_P(\alpha)=\pi (1-\alpha)$ if $\alpha\leq 1$, right? In any case, the point is that I'm only interested in smoothness of $\lambda$ in the interior of its support (hence the requirement $\alpha>0$ in the original question). –  user17240 Jul 5 '12 at 1:30
    
@user17240 Take $P=(x^2+y^2)^{10}-1$: the set $|P|<\alpha$ is the annulus with radii $(1\pm \alpha)^{1/20}$ and the area $\pi(1+ \alpha)^{1/10}-\pi(1- \alpha)^{1/10}$ (the second term disappears when $\alpha>1$). This is not smooth at $\alpha=1$. –  user31373 Jul 5 '12 at 1:33

2 Answers 2

up vote 2 down vote accepted

I think it is indeed $C^\infty$ smooth. Let $r,\theta$ be polar coordinates. For each fixed $\theta$ the function $p_\theta(r)=P(r,\theta)$ is $C^\infty$ smooth on $(0,\infty)$ and has strictly positive derivative. Therefore, the inverse $R_\theta : (0,\infty)\to(0,\infty)$ is also $C^\infty$ smooth. We have $\lambda_P(\alpha)=\frac{1}{2}\int_0^{2\pi}R_\theta^2(\alpha)\,d\theta$, and this can be differentiated with respect to $\alpha$ as many times as we wish.

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Let's modify the problem slightly, considering instead a $1$-variable strictly convex monic polynomial $P(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_0$ and $\lambda_P(\alpha)=|\{x\in\mathbb R: P(x)<\alpha\}|$. Since $P$ is strictly convex, any line intersects it at most twice including the line $y=\alpha$. In fact, it is easy to see that $P$ has even degree and thus $y=\alpha$ intersects $P$ at exactly two points for $\alpha>M$ the global minimum. Let $m$ be the unique value such that $P(m)=M$. Then we have two functions $f_1(\alpha)< m<f_2(\alpha)$ which give the upper and lower roots of $P(x)-\alpha$. Note that $\gamma_P(\alpha)=f_2(\alpha)-f_1(\alpha)$ so if $f_1,f_2$ are smooth then $\lambda_P$ is. Furthermore, since the coefficients $(a_{n-1},\ldots,a_1,a_0-\alpha)$ of $P(x)-\alpha$ are smooth functions of the roots $(f_1,f_2,r_3,\ldots,r_n)$, it follows from the inverse function theorem that the roots are locally given by smooth functions (I think we can get $C^\infty$) of the coefficients wherever the Jacobian is noninvertible. By the implicit function theorem, if the coefficients vary smoothly with several variables we get that $f_1$ and $f_2$ are locally smooth functions of these variables where the Jacobian is noninvertible.

Generalizing the original problem, let $P(x_1,\ldots,x_k)$ be a strictly convex polynomial in arbitrarily many variables. Let $\chi_P(x_1,\ldots,x_k)$ be the indicator function for the set $\{(x_1,\ldots,x_k):P(x_1,\ldots,x_k)<\alpha\}$. By Fubini's theorem, $$\lambda_P(\alpha)=\int_{\mathbb R^n} \chi_P(x_1,\ldots,x_k)dx_1\cdots dx_k=\int_{\mathbb R}\cdots \int_{\mathbb R} \chi_P(x_1,\ldots,x_k)dx_1\cdots dx_k$$ and note that $\chi_P(x_1,\ldots,x_k)$ is the indicator function of a set $\{x_1: f(x_1)<\alpha\}$ where $f$ is a polynomial with coefficients varying smoothly in the other varaibles. It follows that the roots $f_1,f_2$ are smooth functions of $\alpha,x_2,\ldots,x_k$ except where the relevant Jacobian is noninvertible, which by strict convexity corresponds to points at which $f_1,f_2$ fail to exist and so $f_1,f_2$ are smooth wherever they are defined. Since $$\lambda_P(\alpha)=\int_{\mathbb R}\cdots \int_{\mathbb R} (f_2(\alpha,x_2,\ldots,x_n)-f_1(\alpha,x_2,\ldots,x_n)) dx_2\cdots dx_k$$ it follows by differentiating under the integrals that $\lambda_P$ is smooth.

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