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Coercivity makes a small form step a strict contraction
Statement
Assume Countable Choice, used through A bounded form is represented by a unique bounded operator. Let be a real or complex Hilbert space and let be a bounded coercive sesquilinear form on with constants (so necessarily ); let be its operator. Put for . Then , and for every the map is a strict contraction of with constant : for all . In particular is a strict contraction with the same constant. The estimate is the only place where the coercivity constant and the bound enter the contraction argument.
Facts & Assumptions
Given: Countable Choice; a Hilbert space with inner product linear in the first argument; a bounded coercive sesquilinear form with constants ; its operator , ; and a real with .
Inner-product norm expansion: , and by Cauchy--Schwarz; the induced length is a norm with the triangle inequality (Cauchy–Schwarz: , with equality exactly for dependent pairs, The induced length is a norm, Real and imaginary parts, complex conjugation, and modulus).
A map is a strict contraction with constant when for all and (Lipschitz map, -Hölder map for rational , and contraction).
Nonnegative square roots: for there is a unique with , denoted (Square roots exist: a unique with ; the positives are ). If , then would imply , a contradiction; hence .
Proof
On a nonzero Hilbert space the named constants satisfy : choosing and dividing by its norm, , The identity makes the radicand nonnegative; it may vanish when and .
Contraction estimate: for and the expansion of [F2] together with [F1] gives so ; taking and using gives the contraction estimate for every .
The constant lies in : the radicand is a quadratic in with minimum at , which is nonnegative because by step 1.1; at the endpoints and it equals , and for either directly or by the strict minimum unless and , in which case the radicand vanishes and ; in every case . Hence , and with it , is a strict contraction with constant .
Depends on
- Square roots exist: a unique $\sqrt{a} \ge 0$ with $(\sqrt{a})^2 = a$; the positives are $\{x^2 : x \neq 0\}$
- The induced length is a norm
- Bounded, coercive and symmetric sesquilinear forms
- Real and imaginary parts, complex conjugation, and modulus
- Lipschitz map, $\alpha$-Hölder map for rational $0 < \alpha \le 1$, and contraction
- Real powers for positive bases, with the zero-base positive-exponent convention
- A coercive form operator is bounded below
- A bounded form is represented by a unique bounded operator
- Cauchy–Schwarz: $|\langle x,y\rangle|\le\|x\|\,\|y\|$, with equality exactly for dependent pairs
Used by
- The Lax--Milgram theorem Theorem
Dependency tree · two levels
43 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.
Sources
- Haim Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations (Springer Universitext, 2011, complete 614-page text) (standard reference, not scraped)
- John K. Hunter, Notes on Partial Differential Equations (UC Davis, revised 18 June 2014, complete 242-page two-quarter notes) (standard reference, not scraped)
- Richard S. Laugesen, Linear Analysis and Partial Differential Equations (University of Illinois, 2020, complete 158-page graduate notes) (standard reference, not scraped)