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Measure, Integration & Real Analysis, 2021a

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306 Chapter 10 Linear Maps on Hilbert Spaces<br />

By definition, isometries preserve norms. The equivalence of (a) and (b) in the<br />

following result shows that isometries also preserve inner products.<br />

10.60 isometries preserve inner products<br />

Suppose T is a bounded operator on a Hilbert space V. Then the following are<br />

equivalent:<br />

(a) T is an isometry.<br />

(b) 〈Tf, Tg〉 = 〈 f , g〉 for all f , g ∈ V.<br />

(c) T ∗ T = I.<br />

(d) {Te k } k∈Γ is an orthonormal family for every orthonormal family {e k } k∈Γ<br />

in V.<br />

(e) {Te k } k∈Γ is an orthonormal family for some orthonormal basis {e k } k∈Γ<br />

of V.<br />

Proof<br />

If f ∈ V, then<br />

‖Tf‖ 2 −‖f ‖ 2 = 〈Tf, Tf〉−〈f , f 〉 = 〈(T ∗ T − I) f , f 〉.<br />

Thus ‖Tf‖ = ‖ f ‖ for all f ∈ V if and only if the right side of the equation above<br />

is 0 for all f ∈ V. Because T ∗ T − I is self-adjoint, this happens if and only if<br />

T ∗ T − I = 0 (by 10.46). Thus (a) is equivalent to (c).<br />

If T ∗ T = I, then 〈Tf, Tg〉 = 〈T ∗ Tf, g〉 = 〈 f , g〉 for all f , g ∈ V. Thus (c)<br />

implies (b).<br />

Taking g = f in (b), we see that (b) implies (a). Hence we now know that (a), (b),<br />

and (c) are equivalent to each other.<br />

To prove that (b) implies (d), suppose (b) holds. If {e k } k∈Γ is an orthonormal<br />

family in V, then 〈Te j , Te k 〉 = 〈e j , e k 〉 for all j, k ∈ Γ, and thus {Te k } k∈Γ is an<br />

orthonormal family in V. Hence (b) implies (d).<br />

Because V has an orthonormal basis (see 8.67 or 8.75), (d) implies (e).<br />

Finally, suppose (e) holds. Thus {Te k } k∈Γ is an orthonormal family for some<br />

orthonormal basis {e k } k∈Γ of V. Suppose f ∈ V. Then by 8.63(a) we have<br />

f = ∑〈 f , e j 〉e j ,<br />

j∈Γ<br />

which implies that<br />

Thus if k ∈ Γ, then<br />

T ∗ Tf = ∑〈 f , e j 〉T ∗ Te j .<br />

j∈Γ<br />

〈T ∗ Tf, e k 〉 = ∑〈 f , e j 〉〈T ∗ Te j , e k 〉 = ∑〈 f , e j 〉〈Te j , Te k 〉 = 〈 f , e k 〉,<br />

j∈Γ<br />

j∈Γ<br />

where the last equality holds because 〈Te j , Te k 〉 equals 1 if j = k and equals 0<br />

otherwise. Because the equality above holds for every e k in the orthonormal basis<br />

{e k } k∈Γ , we conclude that T ∗ Tf = f . Thus (e) implies (c), completing the proof.<br />

<strong>Measure</strong>, <strong>Integration</strong> & <strong>Real</strong> <strong>Analysis</strong>, by Sheldon Axler

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