L^p norm

Norm from integrating the pth power of absolute value, or essential supremum when p is infinity.
L^p norm

A LpL^p norm on a (X,Σ,μ)(X,\Sigma,\mu) (for 1p<1\le p<\infty) assigns to a f:XRf:X\to\mathbb{R} the value

fp:=(Xf(x)pdμ(x))1/p, \|f\|_p := \left(\int_X |f(x)|^p\,d\mu(x)\right)^{1/p},

whenever the fp|f|^p is finite; here f(x)|f(x)| uses the . For p=p=\infty one defines

f:=ess supxXf(x), \|f\|_\infty := \operatorname*{ess\,sup}_{x\in X} |f(x)|,

using the .

If ff and gg are , then fp=gp\|f\|_p=\|g\|_p, so the LpL^p norm is naturally a norm on the corresponding .

Examples:

  • On ([0,1],B,λ)([0,1],\mathcal{B},\lambda), the constant function f(x)=1f(x)=1 satisfies fp=1\|f\|_p=1 for every 1p1\le p\le\infty.
  • On ([0,1],B,λ)([0,1],\mathcal{B},\lambda), for f(x)=xf(x)=x one has fp=(1/(p+1))1/p\|f\|_p=(1/(p+1))^{1/p} for 1p<1\le p<\infty, and f=1\|f\|_\infty=1.