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The Hilbertian tensor product of H 1 and H 2, sometimes denoted by H 1 ^ H 2, is the Hilbert space obtained by completing H 1 ⊗ H 2 for the metric associated to this inner product. [ 87 ] An example is provided by the Hilbert space L 2 ([0, 1]) .
Note that closed and bounded sets are not in general weakly compact in Hilbert spaces (consider the set consisting of an orthonormal basis in an infinite-dimensional Hilbert space which is closed and bounded but not weakly compact since it doesn't contain 0). However, bounded and weakly closed sets are weakly compact so as a consequence every ...
This function is a test function on and is an element of (). The support of this function is the closed unit disk in . It is non-zero on the open unit disk and it is equal to 0 everywhere outside of it.
The space of all bounded linear operators of finite rank (i.e. that have a finite-dimensional range) is a dense subset of the space of Hilbert–Schmidt operators (with the Hilbert–Schmidt norm). [4] The set of Hilbert–Schmidt operators is closed in the norm topology if, and only if, H is finite-dimensional.
The sesquilinear form B : H × H → is separately uniformly continuous in each of its two arguments and hence can be extended to a separately continuous sesquilinear form on the completion of H; if H is Hausdorff then this completion is a Hilbert space. [1] A Hausdorff pre-Hilbert space that is complete is called a Hilbert space.
In the mathematical discipline of functional analysis, the concept of a compact operator on Hilbert space is an extension of the concept of a matrix acting on a finite-dimensional vector space; in Hilbert space, compact operators are precisely the closure of finite-rank operators (representable by finite-dimensional matrices) in the topology induced by the operator norm.
where H(D) is the space of holomorphic functions in D. Then L 2,h (D) is a Hilbert space: it is a closed linear subspace of L 2 (D), and therefore complete in its own right. This follows from the fundamental estimate, that for a holomorphic square-integrable function ƒ in D
The Hilbert transform of an L 1 function does converge, however, in L 1-weak, and the Hilbert transform is a bounded operator from L 1 to L 1,w. [18] In particular, since the Hilbert transform is also a multiplier operator on L 2 , Marcinkiewicz interpolation and a duality argument furnishes an alternative proof that H is bounded on L p .)
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