Lp 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 subset of $L^1(\mathbb{R})$ is relatively norm compact exactly when it is bounded, its absolute-integral tails tend to zero uniformly, and ${\sup }_{f\in S}\Vert f(\cdot +h)-f{\Vert }_1\to 0$ as $h\to 0$. Necessity follows by taking a finite norm net and using the tail decay and translation continuity in Lp of its finitely many centers; the error contributes at most twice the net radius. For sufficiency, convolve with a triangular unit-integral kernel of width $\delta$. Translation continuity gives uniform approximation in $L^1$, while boundedness gives a common supremum bound and a common Lipschitz constant for the smoothed family. The original tail condition controls its tails outside a slightly enlarged interval. On that interval Arzelà-Ascoli theorem gives a finite uniform net. Extend its centers by zero outside the interval to obtain a finite $L^1$ net, and combine the smoothing and tail errors. Completeness of $L^1$ turns total boundedness into relative compactness.

For a finite interval $(a,b)$ with Lebesgue measure, restrictions of continuous functions on $[a,b]$ form a dense subspace of the Lp space $L^p((a,b))$ for $1\le p<\infty$. One proof first approximates by simple functions. The regularity of Lebesgue measure approximates each of their measurable sets in measure by a finite union of intervals. An interval's indicator function is approximated by continuous functions taking values in $[0,1]$ and differing from that indicator function only in endpoint strips of total length tending to zero. Their Lp norm error is at most the $1/p$ power of that length. The triangle inequality completes the approximation of each simple function.

On $(0,1)$, for $k\ge 0$ and $0\le j<2^k$, the displayed indicator functions form a sequence that converges to zero in every finite Lp norm while failing to converge pointwise anywhere. Indeed, for $1\le p<\infty$,Every point belongs to exactly one interval in each dyadic block, and for $k\ge 1$ it misses at least one other interval in that block. Thus every point sees infinitely many ones and infinitely many zeros. Half-open intervals specify the behaviour at dyadic endpoints. This separates norm convergence in an Lp space from pointwise convergence and convergence almost everywhere.

For each $m\ge 1$, enumerate the $m^2$ half-open intervals $[j/m^2,(j+1)/m^2)$ partitioning $[0,1)$ and use the functions $m$ times their indicator functions. Concatenate these finite blocks in increasing $m$. Every function in block $m$ has L1 norm $1/m$, which tends to zero. Every $x\in (0,1)$ is in one interval of every block, so the values at that point have a subsequence increasing to infinity. Thus small integral norm does not bound the pointwise upper limit, even when every function is nonnegative. In contrast, Fatou's lemma prevents a strictly positive lower limit everywhere on an interval when the norms tend to zero.

The Lp space for $p=1$ consists of equivalence classes of integrable functions, with L1 norm $\Vert f{\Vert }_1=\int |f|d\mu$. Conditional expectation is a contraction in this norm. Convergence in this space is L1 convergence, which controls the change in every integral over a measurable subset.

Suppose measurable $a_n$ on a Lebesgue measurable set $E$ satisfy ${\sum }_n{\int }_E|a_n|<\infty$. The nonnegative partial sums $S_N={\sum }_{n\le N}|a_n|$ increase to a measurable extended-valued $S$. The monotone convergence theorem gives ${\int }_{E}S={\sum }_n{\int }_E|a_n|<\infty$, forcing $S<\infty$ almost everywhere. Each $a_n$ is finite almost everywhere too; discard the union of these countably many null sets when defining signed partial sums. The series ${\sum }_n{a}_n$ then converges absolutely outside a null set.

Suppose ${\sum }_n\Vert a_n{\Vert }_1<\infty$. By summable L1 norms imply almost-everywhere absolute convergence, let $A={\sum }_n{a}_n$ where the series converges absolutely, and choose zero on the exceptional null set. Then $A$ is a Lebesgue measurable function and $|A|\le S={\sum }_n|a_n|$ almost everywhere. Since $S$ is Lebesgue integrable, so is $A$. The signed partial sums are bounded in absolute value by $S$, so the dominated convergence theorem proves termwise integration. It also gives the quantitative L1 norm estimateThe numerical series of integrals is absolutely convergent because $|\int a_n|\le \int |a_n|$.

For example, on $(0,1)$ take $a_n(x)=(-1{)}^{n}n{1}_{(0,2^{-n})}(x)$. Its L1 norm is $n{2}^{-n}$, and these norms sum to two. Every positive $x$ lies in only finitely many supports, so the series is pointwise finite, although its terms are not uniformly small. Termwise integration gives ${\int }_0^1{\sum }_n{a}_n={\sum }_n(-1{)}^{n}n{2}^{-n}=-2/9$, by differentiating the geometric series. This illustrates why a small L1 norm can permit tall terms on short intervals.

The sum consists of measurable functions $f=g+h$ with $g\in L^{p_0}$ and $h\in L^{p_1}$, with norm ${\inf }_{f=g+h}(\Vert g{\Vert }_{p_0}+\Vert h{\Vert }_{p_1})$. It is a Banach space. One construction identifies it with the quotient of $L^{p_0}{\oplus }_1{L}^{p_1}$ by the closed kernel of the addition map. Bounded endpoint operators are therefore bounded on the corresponding sum spaces.

If $p_0<p<p_1$, the displayed split of an $L^p$ function gives $\Vert g{\Vert }_{p_0}\le a^{1-p/p_0}\Vert f{\Vert }_p^{p/p_0}$ and $\Vert h{\Vert }_{p_1}\le a^{1-p/p_1}\Vert f{\Vert }_p^{p/p_1}$. Weighted multiplication with $1/p=(1-\theta )/p_0+\theta /p_1$ cancels the threshold, yielding $\Vert g{\Vert }_{p_0}^{1-\theta }\Vert h{\Vert }_{p_1}^{\theta }\le \Vert f{\Vert }_p$. Choosing $a=\Vert f{\Vert }_p$ also gives the continuous inclusion $\Vert f{\Vert }_{L^{p_0}+L^{p_1}}\le 2\Vert f{\Vert }_p$. The analogous split works if the larger endpoint is infinity.

On $[0,1]$, the continuous functions $f_n(x)=\sin (nx)$ do not form a Cauchy sequence in the Lp space with $p=1$. Put $d_n=f_n-f_{2n}$. The identity ${\int }_0^1{d}_n^{2}dx=1-\sin (2n)/(4n)-\sin (4n)/(8n)-\sin n/n+\sin (3n)/(3n)$ tends to $1$. Since $|d_n|\le 2$, one has ${\int }_0^1|d_n|dx\ge \frac{1}{2}{\int }_0^1{d}_n^{2}dx$, which stays bounded away from zero. Thus arbitrarily late pairs are separated in the integral norm; in particular the sequence cannot converge in that norm.

For left Haar measure, $T_{x}f(t)=f(x^{-1}t)$ is an isometry on $L^p(G)$, $1\le p<\infty$. For compactly supported continuous functions, uniform continuity on a slightly enlarged compact support proves norm continuity. Approximation by those functions and the triangle inequality extend it to all $L^p$. The continuous image of a compact set of translation parameters is consequently compact in $L^p$.

For $1<p<\infty$ and $u,v$ in an Lp space, the derivative with respect to the real scalar $t$ has the displayed form; the integrand is zero where $u=0$. The pointwise chain rule gives the derivative, while the Holder inequality makes $(|u|+|v|{)}^{p-1}|v|$ integrable and permits the dominated convergence theorem. This variational derivative characterizes nearest points to closed convex sets and constructs the Lp duality density of a bounded linear functional.

For $2\le p<\infty$, a nonempty closed convex set $K$ in an Lp space has exactly one nearest point to every $f$. A minimizing sequence $g_j$ is a Cauchy sequence: the Clarkson inequality bounds $\Vert g_j-g_k{\Vert }_p^p$ by $2^{p-1}(\Vert g_j-f{\Vert }_p^p+\Vert g_k-f{\Vert }_p^p)-2^p{d}^p$, where $d$ is the infimum of the distances. Completeness of Lp spaces and the closed set property give a limit in $K$. Applying the same inequality to two minimizers proves uniqueness. Nonemptiness is essential.

From an Lp norm Cauchy sequence choose a subsequence with successive difference norms at most $2^{-j}$. Minkowski inequality and monotone convergence theorem make the sum of the absolute differences an $L^p$ function, hence finite almost everywhere. The telescoping series defines an $L^p$ limit; the geometric bound on its tail proves norm convergence of the subsequence, and the original Cauchy property proves convergence of the whole sequence. No sigma-finiteness assumption is needed.

For $1<q<\infty$, the weak Lq space consists of measurable functions for which $A_q(f)={\sup }_{t>0}t|\{|f|>t\}{|}^{1/q}$ is finite. This distribution-function expression is a quasi-norm. The equivalent norm $N_q(f)={\sup }_{0<|E|<\infty }|E{|}^{-1/q^{\prime }}{\int }_E|f|$ satisfies $A_q\le N_q\le q^{\prime }{A}_q$, with $q^{\prime }=q/(q-1)$. The layer cake representation bounds the integral by ${\int }_0^{\infty }\min (|E|,(A_q/t{)}^q)dt=q^{\prime }{A}_q|E{|}^{1/q^{\prime }}$. The lower bound uses finite-measure subsets of a superlevel set.

The integral expression for the weak Lq space is homogeneous, positive definite up to equality almost everywhere, and obeys the triangle inequality by integrating $|f+g|\le |f|+|g|$ over each set. Its exponent is $-1/q^{\prime }$, not $-1/q$. The sharp comparison constant with the distribution-function quasi-norm is $q^{\prime }$; the power function $x^{-1/q}$ on $(0,\infty )$ attains that ratio on initial intervals.

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.

For a bounded measurable function with positive essential supremum on a finite measure space, put $M_n=\int |f{|}^n$. The Cauchy-Schwarz inequality makes these moments log-convex, so successive ratios are nondecreasing. Their product satisfies $M_n/M_0\le (M_{n+1}/M_n{)}^n$, while each ratio is at most the essential supremum. Positive-measure level sets arbitrarily close to that supremum make $\mu (E{)}^{-1/n}\Vert f{\Vert }_n$ converge to it. Squeezing then proves the ratio limit.

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.

Suppose $1\le p<q<r<\infty$, $\Vert f{\Vert }_p\le C_p$, $\Vert f{\Vert }_q\ge C_q>0$ and $\Vert f{\Vert }_r\le C_r$. Set $\epsilon =(C_q^q/(2C_p^p){)}^{1/(q-p)}$. The part of $\int |f{|}^q$ where $|f|\le \epsilon$ is at most ${\epsilon }^{q-p}{C}_p^p=C_q^q/2$. On the complementary level set, the Holder inequality bounds the integral by $C_r^q\mu \{|f|>\epsilon {\}}^{1-q/r}$, proving the bound. For $r=\infty$ the bound becomes $C_q^q/(2C_r^q)$. The lower-exponent bound prevents spreading into low amplitudes, while the higher-exponent bound prevents concentration into arbitrarily narrow spikes.

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 exponentially weighted Hilbert space consists of measurable functions with ${\int }_{\mathbb{R}}|u(x){|}^2{e}^{2\gamma x}dx<\infty$. Multiplication by $e^{\gamma x}$ is a unitary map onto L2 space. Requiring unweighted square integrability as well produces an intersection that is incomplete under the weighted norm alone. For example, $1_{[-N,0]}$ is a weighted Cauchy sequence for $\gamma >0$, but its weighted limit $1_{(-\infty ,0]}$ is not in unweighted L2 space.

The linearization at a travelling wave $w_c$ is $A={\partial }_z^2+c{\partial }_z+k-2w_c$. Conjugation by $e^{\gamma z}$ changes this to ${\partial }_z^2+(c-2\gamma ){\partial }_z+{\gamma }^2-c\gamma +k-2w_c$. The choice $\gamma =c/2$ removes the first derivative and makes the operator self-adjoint. Its two limiting potentials are $-k-c^2/4$ and $k-c^2/4$, so essential-spectrum invariance under relatively compact perturbations gives ${\sigma }_{\mathrm{ess}}=(-\infty ,k-c^2/4]$. This lies strictly below zero for $c>2\sqrt{k}$ but touches zero at the critical speed. Stability of this essential spectrum alone is not a general nonlinear stability theorem.

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.

For an orthonormal family in a Hilbert space, the squared norm of every finite partial sum is the sum of squared coefficients. Square summability therefore makes the net of finite partial sums Cauchy, proving unconditional convergence by completeness. If the orthonormal family is complete, Bessel's inequality makes the coefficients of each vector square summable, and the residual after summing is orthogonal to the whole family, hence zero. This recovers the orthonormal expansion and Parseval identity.

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.