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Homenormed vector space

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# normed vector space

Let $\mathbb{F}$ be a field which is either $\mathbb{R}$ or $\mathbb{C}$. A *normed vector space* over $\mathbb{F}$ is a pair $(V,\lVert\cdot\rVert)$ where $V$ is a vector space over $\mathbb{F}$ and $\lVert\cdot\rVert\colon V\to\mathbb{R}$ is a function such that

1. $\lVert v\rVert\geq 0$ for all $v\in V$ and $\lVert v\rVert=0$ if and only if $v=0$ in $V$ (

*positive definiteness*)2. $\lVert\lambda v\rVert=\lvert\lambda\rvert\lVert v\rVert$ for all $v\in V$ and all $\lambda\in\mathbb{F}$

3. $\lVert v+w\rVert\leq\lVert v\rVert+\lVert w\rVert$ for all $v,w\in V$ (the

*triangle inequality*)

The function $\lVert\cdot\rVert$ is called a *norm* on $V$.

Some properties of norms:

1. If $W$ is a subspace of $V$ then $W$ can be made into a normed space by simply restricting the norm on $V$ to $W$. This is called the induced norm on $W$.

2. Any normed vector space $(V,\lVert\cdot\rVert)$ is a metric space under the metric $d\colon V\times V\to\mathbb{R}$ given by $d(u,v)=\lVert u-v\rVert$. This is called the

*metric induced by the norm $\lVert\cdot\rVert$*.3. It follows that any normed space is a locally convex topological vector space, in the topology induced by the metric defined above.

4. In this metric, the norm defines a continuous map from $V$ to $\mathbb{R}$ - this is an easy consequence of the triangle inequality.

5. If $(V,\langle,\rangle)$ is an inner product space, then there is a natural induced norm given by $\lVert v\rVert=\sqrt{\langle v,v\rangle}$ for all $v\in V$.

6. The norm is a convex function of its argument.

## Mathematics Subject Classification

46B99*no label found*

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## Attached Articles

equivalent norms by matte

every subspace of a normed space of finite dimension is closed by Mathprof

every finite dimensional normed vector space is a Banach space by matte

extended norm by rspuzio

scaling of the open ball in a normed vector space by matte

Mazur-Ulam theorem by yark

quotient norm by asteroid

## Corrections

norm by matte ✓

Convexity property by Andrea Ambrosio ✓

synonyms and defines by yark ✓

normed vector space by perucho ✓

## Comments

## inner product space?

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.

-apk

## Re: inner product space?

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.

## Re: inner product space?

It seemes so but IÂ´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 canÂ´t succeed thou =( Someone help?

## Re: inner product space?

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. :)

joking

## Re: inner product space?

Thank you m8 that was great :)