Question #219721
What do you mean by Norm of an Inner Product Space
1
Expert's answer
2021-07-22T17:55:44-0400

Any inner product induces a norm given by


v=v,v\|v\|=\sqrt{\langle v, v\rangle}

Proof. The axioms for norms mostly follow directly from those for inner products.

If u,vVu, v ∈ V and αF,α ∈ F, then

(i)

v=v,v0,\|v\|=\sqrt{\langle v, v\rangle}\geq0,

since v,v0\langle v, v\rangle\geq0 with equality if and only if v=0.v = 0.


(ii)

αv=αv,αv=α2v,v\|\alpha v\|=\sqrt{\langle \alpha v,\alpha v\rangle}=\sqrt{|\alpha|^2\langle v, v\rangle}

=αv,v=αav=|\alpha|\sqrt{\langle v, v\rangle}=|\alpha|\|a v\|

(iii) The triangle inequality


Cauchy-Schwarz inequality

If VV is an inner product space, then


u,uuv|\langle u, u\rangle|\leq\|u\|\|v\|

for all u,vV.u, v ∈ V . Equality holds exactly when uu and vv are linearly dependent.


Using the Cauchy-Schwarz inequality,


u+v2=u+v,u+v\|u+ v\|^2=\langle u+v, u+v\rangle

=u,u+u,v+v,u+v,v=\langle u, u\rangle+\langle u, v\rangle+\langle v, u\rangle+\langle v, v\rangle

=u2+u,v+u,v+v2=\|u\|^2+\langle u, v\rangle+\overline{\langle u, v\rangle}+\|v\|^2

=u2+2Reu,v+v2=\|u\|^2+2Re\langle u, v\rangle+\|v\|^2

u2+2u,v+v2\leq\|u\|^2+2|\langle u, v\rangle|+\|v\|^2

u2+2uv+v2\leq\|u\|^2+2\|u\|\|v\|+\|v\|^2

=(u+v)2=(\|u\|+\|v\|)^2

Taking square roots yields


u+vu+v,\|u+ v\|\leq\|u\|+\|v\|,

since both sides are nonnegative.


Therefore v=v,v\|v\|=\sqrt{\langle v, v\rangle} is a Norm.



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS