TheoremBase

Every Linear Operator on a Space with a Finite Orthonormal Basis is Bounded

Statement

Let VV together with ⟨⋅,⋅⟩\langle\cdot,\cdot\rangle be a complex inner product space with induced norm ∥⋅∥\lVert\cdot\rVert, which is a norm on VV by claim 2 of The Induced Norm is a Norm, and Induces a Metric. Let nn be a natural number, let e∈Vne\in V^{n} be an nn-tuple in VV that is an orthonormal basis of VV, with components eke_{k}, and let TT be a linear operator on VV.

Then TT is a bounded linear operator; more precisely, with the finite sum of real numbers

C=∑k=1n∥T(ek)∥,C=\sum_{k=1}^{n}\bigl\lVert T(e_{k})\bigr\rVert ,

one has 0≤C0\le C and

∥T(u)∥≤C ∥u∥for every u∈V,\lVert T(u)\rVert\le C\,\lVert u\rVert\qquad\text{for every }u\in V,

the order being that of the ordered field of real numbers.

Proofs

Log in to submit a proof.

Loading...

Citations

Loading…

Dependencies

Loading…

Related

0 relations

Curated associations between results. These are editable and subjective — they do not replace the dependency graph, which is derived from the references in the text.

No relations recorded yet.

Comments

Log in to comment.

Loading…