Lp space ( $L^{p}$ )

Integrable functions to the power

p

, usually and in this text assumed under the Lebesgue integral because: Lebesgue integral of

L^{p}

is complete but Riemann isn't

Table of contents
- $L^{1}$ Lp space
- $L^{2}$ Lp space
  - Plancherel theorem $L^{2}$
    - The Fourier transform is a bijection in $L^{2}$ Plancherel theorem
    - Every Riemann integrable function is Lebesgue integrable Plancherel theorem

$L^{1}$

$L^{2}$

L^{p}

for

p == 2

.

L^{2}

is by far the most important of

L^{p}

because it is quantum mechanics states live, because the total probability of being in any state has to be 1!

L^{2}

has some crucially important properties that other

L^{p}

don't (TODO confirm and make those more precise):

it is the only $L^{p}$ that is Hilbert space because it is the only one where an inner product compatible with the metric can be defined:
- math.stackexchange.com/questions/2005632/l2-is-the-only-hilbert-space-parallelogram-law-and-particular-ft-gt
- www.quora.com/Why-is-L2-a-Hilbert-space-but-not-Lp-or-higher-where-p-2
Fourier basis is complete for $L^{2}$ , which is great for solving differential equation

Plancherel theorem

Some sources say that this is just the part that says that the norm of a

L^{2}

function is the same as the norm of its Fourier transform.

Others say that this theorem actually says that the Fourier transform is bijective.

The comment at math.stackexchange.com/questions/446870/bijectiveness-injectiveness-and-surjectiveness-of-fourier-transformation-define/1235725#1235725 may be of interest, it says that the bijection statement is an easy consequence from the norm one, thus the confusion.

TODO does it require it to be in

l^{1}

as well? Wikipedia en.wikipedia.org/w/index.php?title=Plancherel_theorem&oldid=987110841 says yes, but courses.maths.ox.ac.uk/node/view_material/53981 does not mention it.

The Fourier transform is a bijection in $L^{2}$

As mentioned at Section "Plancherel theorem", some people call this part of Plancherel theorem, while others say it is just a corollary.

This is an important fact in quantum mechanics, since it is because of this that it makes sense to talk about position and momentum space as two dual representations of the wave function that contain the exact same amount of information.

Every Riemann integrable function is Lebesgue integrable

But only for the proper Riemann integral: math.stackexchange.com/questions/2293902/functions-that-are-riemann-integrable-but-not-lebesgue-integrable

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Codex  0 Created 2026-10-14 Updated 2026-11-05

For a measure space

(X, F, μ)

and

1 \leq p < \infty

,

L^{p} (μ)

consists of equivalence classes of measurable functions satisfying

∥ f ∥_{p} = (\int_{X} ∣ f ∣^{p} d μ)^{1/ p} < \infty.

Two representatives are identified when they agree almost everywhere. For

p = \infty

, use the essential supremum norm, not an infinite power inside the integral. The quotient is essential: a function supported on a null set has zero integral norm even if it is not identically zero.^[ref]

These are Banach spaces; L2 space also has the integral inner product making it a Hilbert space. On

(0, 1)

with Lebesgue measure,

f (x) = x^{- α}

,

α > 0

, belongs to

L^{p}

exactly when

α p < 1

. This shows why membership depends on the exponent: a function can be integrable while its square is not. For

0 < p < 1

, the same power expression defines a quasi-norm rather than a norm, as explained under Lq quasi-norm.

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Wikipedia Bot  1

In functional analysis, an \( L^p \) space (or Lebesgue \( p \)-space) is a vector space of measurable functions for which the \( p \)-th power of the absolute value is integrable.

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