Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 8 / 3 / c / Solution

For every real $s$, the elementary weight estimatefollows from $⟨x+y⟩\le \sqrt{2}⟨x⟩⟨y⟩$, using the reverse rearrangement for negative $s$. If $a$ is smooth its Fourier coefficients decay faster than every power, so $c_j=⟨j{⟩}^{|s|}|\hat{a}(j)|$ is summable. The Fourier convolution formula and Young convolution inequality giveInitially this holds for trigonometric polynomials, and density extends it to every $u\in H_s$, agreeing with distributional multiplication. Hence smooth multiplication preserves every order.

If $s>n/2$, the stronger estimate $⟨m{⟩}^s\le C_s(⟨l{⟩}^s+⟨m-l{⟩}^s)$ givesusing the summability estimate from part (b) with $k=0$. Approximation extends the product, and the continuous embedding identifies it with the pointwise product. Completeness from part (a), this estimate and the constant unit make $H_s$ a commutative Sobolev algebra for $s>n/2$. Rescaling the norm by $C_s^{\prime }$ if necessary gives an equivalent strictly submultiplicative Banach-algebra norm.