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Unisolvent degrees of freedom define a unique reference interpolant
Statement
Let be a reference finite element with , dual shape functions , and bounded extensions of its degrees of freedom to a normed function space containing . Then the operator is the unique linear map satisfying for every , and for every .
Facts & Assumptions
Given: The reference finite-element convention of Reference finite elements and unisolvent degrees of freedom and an input on which every degree of freedom is defined.
The degree-of-freedom map is bijective, and the shape functions satisfy . (Reference finite elements and unisolvent degrees of freedom)
Proof
If , then and the empty sum defines the unique linear map ; all degree-of-freedom identities are vacuous and reproduction holds for the sole element . For , define . The sum is finite and each coefficient is linear in , so is a well-defined linear map into . Applying the th degree of freedom gives by [F1].
If , then both and have the same degree-of-freedom vector, by step 1.1; injectivity of therefore gives . If another has the same degree-of-freedom values, then , and injectivity again gives . This proves uniqueness.
Depends on
Used by
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Sources
- Joachim Schöberl, Numerical Methods for Partial Differential Equations, Chapter 4 (standard reference, not scraped)
- Jean-Luc Guermond, MATH 661 Fall 2025, Chapter 11 (standard reference, not scraped)