If a parameter is generating undesirable boundary terms, try averaging over many choices of that parameter

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Quick description

Many tools in analysis and combinatorics, such as integration by parts, summation by parts, the fundamental theorem of calculus, or telescoping series, can create undesirable "boundary" terms when the integral or sum is cut off. When faced with this problem, one can sometimes solve it by averaging over all choices of the parameter that is generating this boundary, or (equivalently) by choosing the parameter randomly and using the probabilistic method.

Prerequisites

Analysis

Example 1

Problem (Non-endpoint Sobolev inequality) Let $d \geq 1$ , and let $1 < p < q \leq \infty$ be such that $\frac{d}{p}-1 < \frac{d}{q}$ . Establish the Sobolev inequality

$\| f \|_{L^q(\R^d)} \leq C ( \|f\|_{L^p(\R^d)} + \|\nabla f \|_{L^p(\R^d)} )$

for all smooth compactly supported functions $f$ , where $C$ denotes a constant that depends on $d,p,q$ .

Solution The strategy here is to obtain some pointwise bound on $f$ in terms of an integral involving other values of $f$ and $\nabla f$ , so that one can appeal to known inequalities, and specifically Young's inequality. Indeed, if we can get a pointwise bound of the form

$|f(x)| \leq C (|f| * K_1)(x) + C (|\nabla f| * K_2)(x)$ (1)

for two kernels $K_1, K_2$ which are bounded in $L^r(\R^d)$ , where $\frac{1}{p} + \frac{1}{r} = \frac{1}{q}+1$ , then we will be done by Young's inequality and the triangle inequality.

From the fundamental theorem of calculus, we can write

$f(x) = f(x+R \omega) - \int_0^R \omega \cdot \nabla f(x+r\omega)\ dr$

for any $x \in \R^d$ , $\omega \in S^{d-1}$ , and $R>0$ , so by the triangle inequality

$|f(x)| \leq |f(x+R \omega)| + \int_0^R |\nabla f(x+r\omega)|\ dr.$

This does indeed express $f$ in terms of values of $f$ and $\nabla f$ at other places, but it is not of the form (1); the measures that one is convolving by here are far too singular. But we can do better by averaging the parameters. If we first average $\omega$ over all directions in the unit sphere, we obtain

$|f(x)| \leq \int_{S^{d-1}} |f(x+R \omega)|\ d\omega + \int_{S^{d-1}} \int_0^R |\nabla f(x+r\omega)|\ dr d\omega,$

which upon converting from polar coordinates back to Cartesian, becomes

$|f(x)| \leq \int_{S^{d-1}} |f(x+R \omega)|\ d\omega + C \int_{|y| \leq R} \frac{|\nabla f(x-y)|}{|y|^{d-1}}\ dy.$

The second term is in the desired form (1), but the first term is still problematic (one is convolving $f$ here with a singular measure supported on a sphere of radius $R$ , rather than an $L^r$ function to which Young's inequality can be applied). But we can average this problem away, by integrating over all choices of $R$ from $1$ to $2$ (say). To simplify the average we will replace the constraint $|y| \leq R$ by the slightly looser constraint $|y| \leq 2$ , and end up with

$|f(x)| \leq \int_1^2 \int_{S^{d-1}} |f(x+R \omega)|\ d\omega\ dR + C \int_{|y| \leq 2} \frac{|\nabla f(x-y)|}{|y|^{d-1}}\ dy,$

or, upon a further conversion from polar to Cartesian coordinates,

$|f(x)| \leq C \int_{1 \leq |y| \leq 2} |f(x-y)|\ dy + C \int_{|y| \leq 2} \frac{|\nabla f(x-y)|}{|y|^{d-1}}\ dy.$

Thus, one can bound $|f|$ pointwise by the convolution of $|f|$ with $C 1_{1 \leq |y| \leq 2}$ , plus the convolution of $|\nabla f|$ with $C 1_{|y| \leq 2} \frac{1}{|y|^{d-1}}$ . One can check that one now has the desired representation (1).