Lebesgue space

The space $L^p$ consists of measurable functions with finite $p$-norm, identified when they agree almost everywhere. This quotient is necessary because the integral seminorm vanishes on functions supported on null sets.

A mixed Lebesgue norm first takes the Lp norm in one variable and then the norm in another. For finite $p,q$, $\Vert f{\Vert }_{L_x^p{L}_v^q}=(\int (\int |f(x,v){|}^{q}dv{)}^{p/q}dx{)}^{1/p}$; an infinite exponent replaces that integral norm by an essential supremum. The order matters: generally $L_x^p{L}_v^q$ and $L_v^q{L}_x^p$ are different spaces. This distinction is crucial in free transport dispersion.

A measurable function belongs to $L_{\mathrm{loc}}^2(U)$ when its squared absolute value has finite integral on every compact subset of $U$. Functions are identified if they agree almost everywhere. This is a local integrability condition, not a specified global Hilbert space norm. For example, $x$ belongs to $L_{\mathrm{loc}}^2(\mathbb{R})$ but its mean-square averages on expanding intervals diverge.

For $1\le p<\infty$ and $f\in L^p({\mathbb{R}}^d)$, translations converge to $f$ in the Lp space norm. First prove this for continuous compactly supported functions, using uniform continuity and a common bounded support. Their density in Lp space, together with the norm-preserving translation map and the triangle inequality, proves the assertion for arbitrary $f$. It implies convergence of averages over shrinking intervals, a basic approximate identity argument. The corresponding assertion fails for general $L^{\infty }$ functions, such as an indicator function with a jump.

For $f\in L^1(\mathbb{T},m)$ with $m$ normalized Lebesgue measure, translation on the circle group is continuous in the L1 norm: $\Vert f(\cdot +t)-f{\Vert }_1\to 0$ as $t\to 0$ modulo one. Approximate $f$ in the L1 norm by a continuous function, use its uniform continuity, and use translation invariance to preserve the approximation error. In particular, $m(B△(B-t))\to 0$ for every measurable $B$.

On a measure space of finite measure, a bounded measurable function satisfies the displayed limit, where the right side is the essential supremum of its absolute value. The upper estimate is $\Vert f{\Vert }_p\le \mu (X{)}^{1/p}\Vert f{\Vert }_{\infty }$. For every $a<\Vert f{\Vert }_{\infty }$, the set where $|f|>a$ has positive measure, giving the reverse limiting estimate. For a continuous function on an open Euclidean domain, the essential supremum equals the ordinary supremum. Finiteness of the measure is part of this statement.

If $1\le p,r\le \infty$, $0\le \theta \le 1$ and $1/q=\theta /p+(1-\theta )/r$, then the displayed bound follows by applying the Holder inequality to $|u{|}^{\theta q}|u{|}^{(1-\theta )q}$. Interpret the endpoint cases using the essential supremum. In particular convergence in Lp space and boundedness in a larger-exponent Lp space imply convergence at intermediate exponents.

A measurable function is square-integrable when the integral of its squared absolute value is finite. Its equivalence class belongs to the L2 space; on a probability space, this means a finite second moment.

The space $L^2(X,\mathcal{F},\nu )$ consists of almost-everywhere equivalence classes of measurable functions with $\int |f{|}^{2}d\nu <\infty$. The inner productinduces its norm, and the Riesz-Fischer theorem makes it complete; hence it is a Hilbert space.

The mean-zero L2 space is the kernel of the bounded linear functional $h\mapsto Ph$ on L2 space for a probability measure $P$. It is a closed subspace of a Hilbert space, with inner product $P(hk)$, and is the natural ambient space for score functions and canonical gradients.

The L2 inner product is the inner product $⟨f,g⟩=\int f\overline{g}d\mu$ on L2 space. Its induced norm is the L2 norm, and the mass matrix represents it in a finite element basis.

For $1\le p\le \infty$, the Lebesgue space $L^p$ is complete in its usual norm.

The intersection $L^1\cap L^2$ is Banach for $\Vert f{\Vert }_1+\Vert f{\Vert }_2$, but it is generally incomplete when equipped with the $L^1$ norm alone. Truncations of an $L^1$ function outside $L^2$ provide an $L^1$-Cauchy sequence with no limit in the intersection.