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Answer to Question #2884 in Functional Analysis for fairy

Question #2884
Prove every inner product vector space is norm space?
Expert's answer

Let V be a vector space and <,> be an inner product.
This is a function
lt;,>: V x V& --> R
satisfying the following conditions:

1) <a,b>=<b,a>

2) <a,b+c>=<a,b> + <a,c>

3) <ta,b> = t <a,b> for any real t and a,b from V

4) <a,a> is non-negative, moreover,& <a,a>=0 if and only if a=0


Then we can define a norm on X by the following formula:

||v|| = square_root(<v,v>)

Let us verify that ||v|| is a norm indeed.
By definition a norm is a function
||*||: V --> R
such that

A) ||v|| is non-negative and ||v||=0 if and only if v=0

B) ||t v|| = |t|& ||v||, for every real t, and v from V.

C) ||v+w|| <= ||v|| + ||w||


Condition A) follows from 4)

B) ||t v|| = square_root(<tv,tv>) = square_root(t^2 <v,v>) = t square_root(<v,v>) =
t ||v||


Condition C) is a consequence of Cauchy-Bunyakovsky-Schwarz inequality
claiming that
|<v,w>|^2& <=& <v,v> <w,w> = ||v||^2 ||w||^2

Indeed,

||v+w||^2& =& <v+w,v+w>& =& <v,v> + 2<v,w> + <w,w> lt;=
& lt;=& ||v||^2 + 2||v|| ||w|| + ||w||^2& = (||v|| + ||w||)^2.

Hence
||v+w|| <= ||v|| + ||w||

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