Lᵖ Spaces
The Lebesgue integral
does something the Riemann integral never could: it measures the size of a function itself. If
\int |f|\,d\mu is a finite number, that number is a legitimate "length" for
f — and once functions have lengths, they have distances
between them, and for one magic exponent even angles. Functions stop being graphs on a
page and become vectors in a space, one with genuine, if infinite-dimensional, geometry.
That is the single idea of this page: the Lᵖ spaces. They are the vector spaces of
integrable functions, made into normed spaces by the integral. They are complete, so limits
behave. And the case p = 2 is a Hilbert space — the
infinite-dimensional twin of Euclidean space, complete with orthogonality and projections. This is not
an abstraction for its own sake: Lᵖ spaces are the native habitat of Fourier analysis, of partial
differential equations, and of quantum mechanics, where a particle's state is a unit vector in
L^2.
The Lᵖ norm and the space it builds
Fix a measure space (X, \mathcal{M}, \mu) and a real exponent
1 \le p < \infty. For a measurable function f
define its Lᵖ norm as
\|f\|_p \,=\, \left( \int_X |f|^p \, d\mu \right)^{1/p}.
The p-th power inside, the integral, and the p-th root outside are exactly the
ingredients that will make this behave like a length (the root is what earns the triangle inequality
below). The space itself is everything with a finite such length:
L^p(\mu) \,=\, \bigl\{\, f \text{ measurable} : \|f\|_p < \infty \,\bigr\}.
At the other end sits p = \infty. Here the "length" is not an integral but a
ceiling — the smallest bound that holds almost everywhere, the
essential supremum:
\|f\|_\infty \,=\, \operatorname*{ess\,sup}_{x} |f(x)| \,=\, \inf\bigl\{\, M \ge 0 : |f| \le M \text{ a.e.} \,\bigr\}, \qquad L^\infty(\mu) = \{ f : \|f\|_\infty < \infty \}.
The word essential matters: \|f\|_\infty ignores what
f does on any set of measure zero, so a function that is
10^{9} at a single point but bounded by 3 elsewhere
has \|f\|_\infty = 3.
\|\cdot\|_p is almost a norm — but it fails one axiom on the nose.
Recall the vanishing property of the integral:
- \|f\|_p = 0 \iff \int |f|^p\,d\mu = 0 \iff f = 0 almost everywhere — not everywhere.
- So the honest villain \mathbf{1}_{\mathbb{Q}} has \|\mathbf{1}_{\mathbb{Q}}\|_p = 0 yet is not the zero function. On raw functions \|\cdot\|_p is only a seminorm.
- The fix is to declare f \sim g when f = g a.e., and let L^p be the space of these equivalence classes. Then \|f\|_p = 0 means f is the zero class, and \|\cdot\|_p becomes a genuine norm.
This is a subtlety worth internalising early: a point of L^p is a whole
family of functions agreeing a.e., glued together. We keep writing "f \in L^p"
and calling it a function out of habit, but the object that actually lives in the space is its class.
The two inequalities that make it a space
Two theorems do all the heavy lifting. The first controls products, the second controls sums — and it
is the second that certifies \|\cdot\|_p as a norm.
Let p, q \in (1, \infty) be conjugate exponents, meaning
\frac{1}{p} + \frac{1}{q} = 1.
Then for measurable f, g,
\|fg\|_1 = \int |fg|\,d\mu \;\le\; \|f\|_p\,\|g\|_q.
- The proof rests on Young's inequality ab \le \tfrac{a^p}{p} + \tfrac{b^q}{q} (itself just convexity of the exponential), applied pointwise to the normalised functions and integrated.
- The symmetric case p = q = 2 is exactly the Cauchy–Schwarz inequality |\int f\bar g| \le \|f\|_2\,\|g\|_2.
For 1 \le p \le \infty and f, g \in L^p,
\|f + g\|_p \;\le\; \|f\|_p + \|g\|_p.
This is the triangle inequality for \|\cdot\|_p. With it,
the three norm axioms all hold — positivity (on a.e.-classes),
\|cf\|_p = |c|\,\|f\|_p, and the triangle inequality — so
(L^p, \|\cdot\|_p) is a genuine normed vector space, and
d(f, g) = \|f - g\|_p is a genuine metric.
Minkowski for p > 1 is proved using Hölder: split
|f+g|^p \le |f|\,|f+g|^{p-1} + |g|\,|f+g|^{p-1}, integrate, and apply Hölder to
each term with exponent q = p/(p-1) on the trailing factor. The two theorems
are a matched pair.
What does the "unit ball" look like?
A norm is fully pictured by its unit ball — the set of vectors of length
\le 1. In the two-dimensional model space
\mathbb{R}^2 with the p-norm
\|(x,y)\|_p = (|x|^p + |y|^p)^{1/p}, the unit ball is
\{|x|^p + |y|^p \le 1\}, and it changes shape dramatically with
p. Drag the slider and watch:
At p = 1 the ball is a diamond (the taxicab metric); at
p = 2 it is a perfectly round circle; as
p \to \infty it swells into a square (the max-norm). The
roundness at p = 2 is not cosmetic — it is the visible signature of the inner
product, the one exponent where the norm comes from a dot product and rotations are symmetries. The same
story plays out, invisibly, in the infinite-dimensional function spaces L^p:
only L^2 has a "round" geometry with angles.
Completeness: Riesz–Fischer, and why the Lebesgue integral pays off
A normed space is only as useful as its limits. The decisive fact about
L^p is that it is complete: every Cauchy sequence converges
to a limit that is still in the space. A complete normed space is a
Banach space.
For every 1 \le p \le \infty, the space
L^p(\mu) is complete. That is:
- if \|f_n - f_m\|_p \to 0 as n, m \to \infty (a Cauchy sequence),
- then there is an f \in L^p with \|f_n - f\|_p \to 0.
- Hence every L^p is a Banach space, and L^2 is a Hilbert space.
This is exactly where the long climb through measure theory earns its keep. The naive alternative — the
space of Riemann-integrable functions under the same \|\cdot\|_2 — is
not complete: one can build a Cauchy sequence of nice Riemann-integrable functions whose
limit is something like \mathbf{1}_{\mathbb{Q}}, which escapes the space
entirely. Lebesgue's integral is precisely the completion that plugs those holes. Completeness is not a
technical footnote; it is why L^2 works — why Fourier series
converge to genuine functions, why the fixed-point and projection theorems of analysis apply, why the
whole edifice stands.
L² is special: the one with angles
Among all the L^p, the case p = 2 is set apart,
because its norm is born from an
inner product:
\langle f, g \rangle \,=\, \int_X f \, \overline{g} \, d\mu, \qquad\text{so that}\qquad \|f\|_2 = \sqrt{\langle f, f \rangle} = \left(\int |f|^2\,d\mu\right)^{1/2}.
(The complex conjugate \overline{g} is there so
\langle f, f\rangle = \int |f|^2 \ge 0 for complex-valued functions; for real
functions it is just \int fg.) An inner product is far more than a norm — it
encodes angles. Two functions are orthogonal when
\langle f, g\rangle = 0, and with orthogonality come the full toolkit of
Euclidean geometry in infinitely many dimensions: the Pythagorean theorem, orthogonal projection onto a
subspace, and orthonormal bases.
In short: L^2 is a complete inner-product space, i.e. a
Hilbert space — the infinite-dimensional analogue of Euclidean
\mathbb{R}^n. The Cauchy–Schwarz inequality above is just Hölder at
p = q = 2, and it is what lets us define
\cos\theta = \tfrac{\langle f, g\rangle}{\|f\|_2\|g\|_2}. For
p \ne 2 there is no such inner product and no notion of angle at all — a
L^p for p \ne 2 is a Banach space but never a
Hilbert space.
Worked examples: who is in, who is out
Example 1 — a power near the origin: f(x) = x^{-a} on (0,1)
Take a > 0 and ask when f(x) = x^{-a} lies in
L^p(0,1). Compute the defining integral directly:
\int_0^1 |x^{-a}|^p \, dx = \int_0^1 x^{-ap}\, dx, \qquad\text{which is finite} \iff ap < 1 \iff a < \frac1p.
The singularity at 0 is only integrable if the exponent
ap stays below 1. So
x^{-1/2} \in L^1(0,1) but x^{-1/2} \notin L^2(0,1)
(there ap = 1, borderline divergent). Larger
p punishes the spike more.
Example 2 — the same power far out: f(x) = x^{-a} on (1,\infty)
Now the trouble is at infinity, and the condition flips:
\int_1^\infty x^{-ap}\, dx \;\text{ is finite} \iff ap > 1 \iff a > \frac1p.
A fat tail needs ap above 1 to decay fast
enough. The two examples together are the classic warning: on the infinite line
\mathbb{R} a single power can be integrable at neither end, and no
inclusion L^p \subseteq L^q holds in general.
Example 3 — nesting on a finite measure space
When the whole space has finite measure — say [0,1] with
\mu([0,1]) = 1 — the powers do nest, and the bigger exponent wins:
\mu(X) < \infty \ \text{ and } \ p_1 > p_2 \quad\Longrightarrow\quad L^{p_1}(\mu) \subseteq L^{p_2}(\mu).
(One line of Hölder: apply it to |f|^{p_2} \cdot 1 with exponent
p_1/p_2, and the finiteness of \mu(X) supplies the
needed factor.) So on [0,1], L^\infty \subset \cdots \subset L^2
\subset L^1 — every bounded function is square-integrable, every square-integrable
function is integrable.
Example 4 — a norm you can compute
Let f be the two-step simple function on [0,1]
equal to 3 on [0,\tfrac12) and
1 on [\tfrac12, 1]. Then
\|f\|_2 = \left(\int_0^1 |f|^2\,dx\right)^{1/2} = \left(3^2\cdot\tfrac12 + 1^2\cdot\tfrac12\right)^{1/2} = \sqrt{\tfrac{9}{2} + \tfrac12} = \sqrt{5}.
Its length as a vector in L^2[0,1] is \sqrt5 — a
concrete number attached to a function, computed by exactly the same value-times-measure bookkeeping the
Lebesgue integral gave us.
Yes — that is the whole secret. On [-\pi, \pi] the functions
e_n(x) = \tfrac{1}{\sqrt{2\pi}}\,e^{inx}, for
n \in \mathbb{Z}, are orthonormal in
L^2: \langle e_n, e_m\rangle = \int_{-\pi}^{\pi} e_n \overline{e_m} = \delta_{nm},
a clean 1 when n = m and
0 otherwise. And they are complete: they span the whole space. So a
function's Fourier series is nothing more exotic than its expansion in an orthonormal basis,
f = \sum_{n} \langle f, e_n\rangle \, e_n, \qquad \|f\|_2^2 = \sum_n |\langle f, e_n\rangle|^2,
with that second identity — Parseval's theorem — being the infinite-dimensional
Pythagoras. The convergence \sum \to f is convergence in
L^2, and it holds for every
f \in L^2 precisely because the space is complete (Riesz–Fischer). The
nineteenth century's mess over "does the Fourier series converge?" dissolves once you see it as a
coordinate expansion in a Hilbert space.
Five traps at the door of Lᵖ:
-
Elements of L^p are equivalence CLASSES, not functions.
Asking for "the value of f \in L^p at the point
x = 0" is meaningless: a single point is a null set, so you can
change that value freely without leaving the class. Pointwise value at a point is simply not defined
for an L^p element.
-
\|f\|_p = 0 means f = 0 almost
everywhere, not everywhere. It is the passage to a.e.-classes that turns the seminorm into a
norm; skip that step and the "norm" fails its own positivity axiom.
-
Only p = 2 is an inner-product space. A general
L^p is a Banach space with a distance but no angles — there is no
\langle f, g\rangle giving rise to \|\cdot\|_p
unless p = 2. (The parallelogram law fails for
p \ne 2, which is exactly the obstruction.)
-
Inclusions L^p \subseteq L^q depend on the measure. On a
finite measure space the larger exponent's space is the smaller
(L^{p_1}\subseteq L^{p_2} for p_1 > p_2). On all
of \mathbb{R} (infinite measure) neither contains the other
— Example 1 and 2 above are the counterexamples.
-
Conjugate exponents satisfy \tfrac1p + \tfrac1q = 1, not
p + q anything. So p = 2 pairs with
q = 2, p = 3 with
q = \tfrac32, and p = 1 pairs with
q = \infty at the extreme.