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# $L^{p}$-space

# Definition

Let $(X,\mathfrak{B},\mu)$ be a measure space. Let $0<p<\infty$. The *$L^{{p}}$-norm* of a function $f:X\rightarrow\mathbb{C}$ is defined as

$\left|\left|f\right|\right|_{{p}}:=\left(\int_{{X}}\left|f\right|^{p}d\mu% \right)^{{\frac{1}{p}}}$ | (1) |

when the integral exists. The set of functions with finite $L^{{p}}$-norm forms a vector space $V$ with the usual pointwise addition and scalar multiplication of functions. In particular, the set of functions with zero $L^{{p}}$-norm form a linear subspace of $V$, which for this article will be called $K$. We are then interested in the quotient space $V/K$, which consists of complex functions on $X$ with finite $L^{{p}}$-norm, identified up to equivalence almost everywhere. This quotient space is the complex $L^{{p}}$-space on $X$.

# Theorem

If $1\leq p<\infty$, the vector space $V/K$ is complete with respect to the $L^{{p}}$ norm.

# The space $L^{{\infty}}$.

The space $L^{{\infty}}$ is somewhat special, and may be defined without explicit reference to an integral. First, the $L^{{\infty}}$-norm of $f$ is defined to be the essential supremum of $\left|f\right|$:

$\left|\left|f\right|\right|_{{\infty}}:=\mathrm{ess\ sup}\left|f\right|=\inf% \left\{a\in\mathbb{R}:\mu(\left\{x:\left|f(x)\right|>a\right\})=0\right\}$ | (2) |

However, if $\mu$ is the trivial measure, then essential supremum of every measurable function is defined to be 0.

The definitions of $V$, $K$, and $L^{{\infty}}$ then proceed as above, and again we have that $L^{\infty}$ is complete. Functions in $L^{{\infty}}$ are also called essentially bounded.

# Example

Let $X=[0,1]$ and $f(x)=\frac{1}{\sqrt{x}}$. Then $f\in L^{{1}}(X)$ but $f\notin L^{{2}}(X)$.

## Mathematics Subject Classification

28B15*no label found*

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## Attached Articles

proof that $L^p$ spaces are complete by Simone

$\ell^p(X)$ space by asteroid

bounded linear functionals on $L^p(\mu)$ by azdbacks4234

bounded linear functionals on $L^\infty(\mu)$ by gel

$L^p$-norm is dual to $L^q$ by gel

compactly supported continuous functions are dense in $L^p$ by asteroid

## Comments

## L^\infty special?

I have a question... can't you just define L^\infty as a limit (in p)?

-apk

## Re: L^\infty special?

Won't work for spaces of infinite measure, that

is mu(X)=\infty

For such a space the constant functions are in

L^\infty, but not in any of the L^p.

Doesn't work for spaces of finite measure either

Consider X=[0,1] and f(x) = x

We have ||f||_p = (p+1)^(-p) -> 0

However ||f||_oo = 1

## Re: L^\infty special?

Your example fails : For X=[0,1] and f(x)=x, we have ||f||_p=(p+1)^(-1/p) -> 1 (not (p+1)^(-p)). In fact, the following is true :

THM : For any measure space X and function f in L^\r for some r<\infty, we have lim_{p->\infty}||f||_p=||f||_\infty.

The proof is a simple exercise with Chebyshev's inequality; see "Lebesgue Integration on Euclidean Space" by Frank Jones, p. 239 for more details.

## Re: L^\infty special?

Thanks. My example does fail - I erred. Thanks also for the THM.

## Re: L^\infty special?

I'm not sure that theorem is entirely true either. Take X to be the

degenerate measure space i.e. any subset including itself has measure 0.

and take f=0 which is in L^r for r<00, but

lim_{p->\infty}||0||_p = 0, and ||0||_\infty = -infinity, by the way

the essential supremum is defined.