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normed vector space

norm, metric induced by a norm, metric induced by the norm, induced norm
normed space, normed linear space
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Mathematics Subject Classification

46B99 no label found


Is an inner product space automatically a normed vector space (with ||x||=\sqrt{<x,x>})?

It seems so, since an inner product space taken with the norm defined as above has all of the normed vector space properties, *plus* the Cauchy-Schwartz inequality.

Yes, this is discussed in the entry "inner
product space". The only requirement on the
inner product is for it to be non-degenerate.
That is <a,a> = 0 iff a=0.

It seemes so but I´m trying to prove the triangle inequality right now (the first two are trivial) and all I know is that I have to use the Cauchy-Schwartz inequality but I can´t succeed thou =( Someone help?

Let w,v be vectors from inner product space. Calculate ||w+v||^2 as follows:

||w+v||^2 = <w+v,w+v> = <w,w>+<w,v>+S(<w,v>)+<v,v>

due to linearity (where S(.) denotes complex conjugate ).

Now <w,v>+S(<w,v>) = 2*Re(<w,v>) <= 2*|<w,v>| <= 2*||w||*||v||

due to Cauchy-Schwartz inequality, so:

||w+v||^2 <= ||w||^2 + 2*||w||*||v|| + ||v||^2 = ( ||w|| + ||v|| )^2

so finaly

||w+v|| <= ||w|| + ||v||.

I hope everything's fine. :)


Thank you m8 that was great :)

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