Lp Spaces

Let $(\Omega, \mathcal{F}, \mu)$ be a measure space, $1 \leq p < \infty$, and $f$ a measurable function $\Omega \to \mathbb{R}$. The $L^p$ seminorm of $f$ is

$\|f\|_p \;=\; \left( \int |f|^p \, d\mu \right)^{1/p}.$

This is a seminorm, not a norm: $\|f\|_p \geq 0$, scalar multiplication pulls through ($\|\alpha f\|_p = |\alpha| \, \|f\|_p$), and the triangle inequality holds (Minkowski, Section 5). What it does not satisfy is the definiteness axiom: $\|f\|_p = 0$ does not imply $f = 0$ — only $f = 0$ almost everywhere.

Define the equivalence relation $f \sim g$ on measurable functions by

$f \sim g \;\iff\; f = g \text{ $\mu$-almost everywhere}.$

The space $L^p(\mu)$ is the set of equivalence classes $[f]$ of measurable functions for which $\|f\|_p < \infty$. On $L^p(\mu)$, the seminorm becomes a norm:

$\|[f]\|_p \;=\; 0 \;\iff\; [f] = [0],$

because $\|f\|_p = 0$ now means ”$f = 0$ a.e.”, which is exactly the statement that $[f]$ is the zero equivalence class.

For a measurable function $f$, the essential supremum is

$\operatorname{ess\,sup} |f| \;=\; \inf \bigl\{ M \geq 0 \;:\; |f(x)| \leq M \text{ for $\mu$-a.e. } x \bigr\}.$

The space $L^\infty(\mu)$ consists of equivalence classes (under a.e.-equality) of measurable functions $f$ for which $\operatorname{ess\,sup} |f| < \infty$. Its norm is

$\|f\|_\infty \;=\; \operatorname{ess\,sup} |f|.$

Consider $f(x) = x^{-1/3}$ on $(0, 1]$ with Lebesgue measure. We compute

$\|f\|_p^p \;=\; \int_0^1 x^{-p/3} \, dx \;=\; \begin{cases} \dfrac{1}{1 - p/3} & \text{if } p < 3, \\ \infty & \text{if } p \geq 3. \end{cases}$

So $f \in L^p((0, 1])$ for every $p < 3$, with $\|f\|_p$ growing as $p$ approaches $3$. At $p = 3$ the integral $\int_0^1 x^{-1} \, dx$ diverges logarithmically and the norm becomes infinite, so $f \notin L^3$. The number $3$ is called the critical exponent for this function: for any $p$ below it the function is integrable to the $p$-th power, and for any $p$ at or above it, it is not. Different functions have different critical exponents, and that variability is what makes the family of $L^p$ spaces interesting — the same function can be a member of one space and not another.

Let $f(x) = (\log(1/x))^{-1}$ on $(0, 1/2)$ extended by zero outside. As $x \to 0^+$, $\log(1/x) \to \infty$, so $f(x) \to 0$ — the function is bounded near $0$. But near $x = 1/2$ from below, $\log(1/x) \to \log 2 > 0$, so $f$ approaches a finite positive value. The essential supremum is finite on most of the interval but blows up on the upper boundary in a different way:

$\sup_{x \in (0, 1/2)} f(x) \;=\; \lim_{x \to 1/2^-} \frac{1}{\log(1/x)} \;=\; \frac{1}{\log 2} \;\approx\; 1.44.$

So this particular $f$ is in fact in $L^\infty$. To get a function in every finite $L^p$ but not in $L^\infty$, take instead $f(x) = (\log(1/x))^{1/2}$ on $(0, 1)$: the integrals $\int_0^1 (\log(1/x))^{p/2} \, dx$ are all finite (the logarithm grows slower than any power, so integrating against it converges for every $p$), but $f(x) \to \infty$ as $x \to 0$, so $\operatorname{ess\,sup} |f| = \infty$ and $f \notin L^\infty$. The membership patterns of $L^p$ as $p$ varies are surprisingly subtle, and they reflect different ways a function can fail to be “small.”

We will write $f \in L^p$ throughout this topic when we technically mean $[f] \in L^p$. The abuse is universal in analysis and rarely causes confusion: when a statement holds for $f$, it holds for every representative of $[f]$, since the statements are themselves only sensitive to a.e.-behavior. Two functions in the same equivalence class are interchangeable for every purpose we care about — integrals against them, $L^p$ norms, convergence in $L^p$ — so working with representatives instead of classes is harmless and notationally cleaner.

The quotient by a.e.-equivalence is not optional. Without it, the seminorm $\|\cdot\|_p$ is not a norm, the functional $f \mapsto \|f\|_p$ does not separate points, and the metric $d(f, g) = \|f - g\|_p$ is not a metric — two distinct functions can be at distance zero. None of the geometric constructions we want — open balls, convergence in norm, completeness, orthogonal projection — make sense in a seminormed space, because the topology is not Hausdorff. The quotient promotes the seminorm to a norm by collapsing every equivalence class to a single point, and that single move makes the entire theory of $L^p$ spaces possible.

Let $(\Omega, \mathcal{F}, \mu)$ be a probability space ($\mu(\Omega) = 1$), let $f \in L^1(\mu)$, and let $\phi: \mathbb{R} \to \mathbb{R}$ be convex. Then

$\phi\!\left( \int f \, d\mu \right) \;\leq\; \int \phi \circ f \, d\mu.$

The proof uses one fact about convex functions: at every point, a convex function has a supporting line. That is, for every $a \in \mathbb{R}$ there exists a slope $\lambda \in \mathbb{R}$ (the right or left derivative of $\phi$ at $a$, or any value in between if $\phi$ is not differentiable at $a$) such that

$\phi(t) \;\geq\; \phi(a) + \lambda (t - a) \quad \text{for all } t \in \mathbb{R}.$

This is the definition of $\phi$ being convex, restated geometrically: the graph of $\phi$ lies above every one of its tangent lines.

Step 1: Choose $a$. Set $a = \int f \, d\mu$ — the integral of $f$ against $\mu$. Since $\mu$ is a probability measure and $f \in L^1$, this integral is a finite real number. Choose any supporting line of $\phi$ at $a$ with slope $\lambda$.

Step 2: Apply the supporting-line inequality pointwise. For each $x \in \Omega$ we have $f(x) \in \mathbb{R}$, so the supporting-line inequality with $t = f(x)$ gives

$\phi(f(x)) \;\geq\; \phi(a) + \lambda (f(x) - a).$

This holds at every point $x$, with the same constants $\phi(a)$ and $\lambda$.

Step 3: Integrate against $\mu$. Both sides of the pointwise inequality are measurable functions of $x$, and the right-hand side is integrable because it is an affine function of $f$ and $f \in L^1$. Integrating preserves the inequality:

$\int \phi(f(x)) \, d\mu(x) \;\geq\; \int \bigl[ \phi(a) + \lambda (f(x) - a) \bigr] \, d\mu(x).$

The right side splits using linearity of the integral:

$\int \phi(f) \, d\mu \;\geq\; \phi(a) \cdot \mu(\Omega) + \lambda \int (f - a) \, d\mu \;=\; \phi(a) + \lambda \cdot 0 \;=\; \phi(a).$

We used $\mu(\Omega) = 1$ (probability measure) for the first term and $\int (f - a) \, d\mu = \int f \, d\mu - a = a - a = 0$ for the second. Substituting $a = \int f \, d\mu$ on the right gives the conclusion.

Let $P$ and $Q$ be probability measures on $\Omega$ with $P \ll Q$ (i.e., $P$ is absolutely continuous with respect to $Q$ — every $Q$-null set is $P$-null), and let $g = dP/dQ$ be the Radon–Nikodym density (we will formalize Radon–Nikodym in Topic 28; for now treat $g$ as the “ratio” of $P$ to $Q$). The Kullback–Leibler divergence of $P$ from $Q$ is

$D_{\mathrm{KL}}(P \,\|\, Q) \;=\; \int \log g \, dP \;=\; -\int \log(1/g) \, dP.$

The function $\phi(t) = -\log t$ is convex on $(0, \infty)$. Apply Jensen’s inequality to $f = 1/g$ against the probability measure $P$:

$-\log\!\left( \int (1/g) \, dP \right) \;\leq\; \int -\log(1/g) \, dP \;=\; D_{\mathrm{KL}}(P \,\|\, Q).$

Now compute the integral on the left. Since $g = dP/dQ$, we have $\int (1/g) \, dP = \int (1/g) \cdot g \, dQ = \int 1 \, dQ = 1$. So $-\log(1) = 0$, and we get

$0 \;\leq\; D_{\mathrm{KL}}(P \,\|\, Q),$

with equality iff $1/g$ is constant $P$-a.e. (the equality case of Jensen for strictly convex $\phi$), which means $g$ is constant, which means $g = 1$ a.e., which means $P = Q$. This is Gibbs’ inequality, the foundational non-negativity result of information theory.

Jensen’s inequality is the bridge from measure theory to information theory, and it is the only bridge most people remember after they leave grad school. Every non-negativity statement in information theory — Gibbs’ inequality (KL $\geq 0$), the chain rule for KL divergence, the data processing inequality, the convexity of mutual information — is a Jensen application. Variational methods in machine learning rest on the same foundation: the ELBO (evidence lower bound) used in variational autoencoders and Bayesian neural networks is Jensen applied to $\log p(x) = \log \int p(x, z) \, dz$, and the inequality slack is exactly the KL divergence between the variational posterior and the true posterior. Forward link: Variational Inference.

For $a, b \geq 0$ and conjugate exponents $p, q > 1$ satisfying $1/p + 1/q = 1$:

$ab \;\leq\; \frac{a^p}{p} + \frac{b^q}{q}.$

Equality holds iff $a^p = b^q$.

The cases $a = 0$ or $b = 0$ are trivial, so assume $a, b > 0$.

Step 1: Rewrite the product as an exponential. Take logarithms inside an exponential:

$ab \;=\; \exp(\log a + \log b) \;=\; \exp\!\left( \frac{1}{p} \cdot p \log a + \frac{1}{q} \cdot q \log b \right).$

The exponent on the right is a convex combination of $p \log a$ and $q \log b$, with weights $1/p$ and $1/q$ that sum to $1$ by the conjugacy condition.

Step 2: Apply convexity of the exponential. The function $\phi(t) = e^t$ is convex. For any convex combination $\alpha u + \beta v$ with $\alpha, \beta \geq 0$ and $\alpha + \beta = 1$, convexity gives $e^{\alpha u + \beta v} \leq \alpha e^u + \beta e^v$. With $\alpha = 1/p$, $\beta = 1/q$, $u = p \log a$, and $v = q \log b$:

$\exp\!\left( \frac{1}{p} \cdot p \log a + \frac{1}{q} \cdot q \log b \right) \;\leq\; \frac{1}{p} \exp(p \log a) + \frac{1}{q} \exp(q \log b).$

Step 3: Simplify. Note that $\exp(p \log a) = a^p$ and $\exp(q \log b) = b^q$. Combining the previous two displays:

$ab \;\leq\; \frac{a^p}{p} + \frac{b^q}{q}.$

Equality in the convexity step requires $p \log a = q \log b$ (the two points being averaged are the same), which after exponentiation is $a^p = b^q$.

Let $f \in L^p(\mu)$ and $g \in L^q(\mu)$ with $1 \leq p \leq \infty$ and $1/p + 1/q = 1$ (with the convention $1/\infty = 0$, so $p = 1$ pairs with $q = \infty$ and vice versa). Then $fg \in L^1(\mu)$ and

$\|fg\|_1 \;\leq\; \|f\|_p \, \|g\|_q.$

We prove the case $1 < p < \infty$ (so $1 < q < \infty$ as well). The endpoint cases $p = 1, q = \infty$ and $p = \infty, q = 1$ are immediate from the definition of $\|g\|_\infty$ as the essential supremum: $|fg| \leq \|g\|_\infty \cdot |f|$ a.e., so $\int |fg| \leq \|g\|_\infty \int |f| = \|f\|_1 \|g\|_\infty$.

Step 1: Trivial cases. If $\|f\|_p = 0$, then $f = 0$ a.e., so $fg = 0$ a.e., and both sides of the inequality are zero. Similarly for $\|g\|_q = 0$. So we may assume $\|f\|_p > 0$ and $\|g\|_q > 0$.

Step 2: Normalize. Define the rescaled functions

$\tilde f \;=\; \frac{f}{\|f\|_p}, \qquad \tilde g \;=\; \frac{g}{\|g\|_q}.$

By construction $\|\tilde f\|_p = 1$ and $\|\tilde g\|_q = 1$. Proving $\|\tilde f \tilde g\|_1 \leq 1$ for the normalized pair will give $\|fg\|_1 \leq \|f\|_p \|g\|_q$ after multiplying through by $\|f\|_p \|g\|_q$.

Step 3: Apply Young’s inequality pointwise. For each $x$, set $a = |\tilde f(x)|$ and $b = |\tilde g(x)|$ in Young’s inequality:

$|\tilde f(x) \tilde g(x)| \;=\; |\tilde f(x)| \cdot |\tilde g(x)| \;\leq\; \frac{|\tilde f(x)|^p}{p} + \frac{|\tilde g(x)|^q}{q}.$

This holds at every point $x$, with the same $p$ and $q$.

Step 4: Integrate against $\mu$. Both sides are measurable functions of $x$, and the right-hand side is integrable since $\tilde f \in L^p$ and $\tilde g \in L^q$. Integration preserves the inequality:

$\int |\tilde f \tilde g| \, d\mu \;\leq\; \frac{1}{p} \int |\tilde f|^p \, d\mu + \frac{1}{q} \int |\tilde g|^q \, d\mu.$

By construction, $\int |\tilde f|^p \, d\mu = \|\tilde f\|_p^p = 1^p = 1$, and likewise $\int |\tilde g|^q \, d\mu = 1$. The right side becomes $1/p + 1/q = 1$, so

$\|\tilde f \tilde g\|_1 \;=\; \int |\tilde f \tilde g| \, d\mu \;\leq\; 1.$

Step 5: Un-normalize. Multiplying both sides by $\|f\|_p \|g\|_q$:

$\|fg\|_1 \;=\; \|f\|_p \|g\|_q \cdot \|\tilde f \tilde g\|_1 \;\leq\; \|f\|_p \|g\|_q.$

The most familiar special case of Hölder is the Cauchy–Schwarz inequality:

$\int |fg| \, d\mu \;\leq\; \left( \int |f|^2 \, d\mu \right)^{1/2} \left( \int |g|^2 \, d\mu \right)^{1/2}.$

This is Hölder’s inequality with $p = q = 2$ — and the conjugacy condition $1/2 + 1/2 = 1$ checks out. Cauchy–Schwarz is the inequality that makes $L^2$ an inner product space: if we define $\langle f, g \rangle = \int fg \, d\mu$, then Cauchy–Schwarz says $|\langle f, g \rangle| \leq \|f\|_2 \|g\|_2$, which is the defining property of an inner product. Section 9 develops this perspective in detail.

Consider $f(x) = x^{1/3}$ and $g(x) = x^{1/6}$ on $[0, 1]$ with Lebesgue measure, $p = 3$ and $q = 3/2$ (note $1/3 + 2/3 = 1$). Compute each piece of Hölder:

$\|fg\|_1 \;=\; \int_0^1 x^{1/3} \cdot x^{1/6} \, dx \;=\; \int_0^1 x^{1/2} \, dx \;=\; \frac{2}{3} \;\approx\; 0.6667.$

$\|f\|_3 \;=\; \left( \int_0^1 (x^{1/3})^3 \, dx \right)^{1/3} \;=\; \left( \int_0^1 x \, dx \right)^{1/3} \;=\; \left( \frac{1}{2} \right)^{1/3} \;\approx\; 0.7937.$

$\|g\|_{3/2} \;=\; \left( \int_0^1 (x^{1/6})^{3/2} \, dx \right)^{2/3} \;=\; \left( \int_0^1 x^{1/4} \, dx \right)^{2/3} \;=\; \left( \frac{4}{5} \right)^{2/3} \;\approx\; 0.8618.$

The product $\|f\|_3 \cdot \|g\|_{3/2} \approx 0.7937 \cdot 0.8618 \approx 0.6840$, so Hölder gives

$0.6667 \;\leq\; 0.6840.$

The inequality holds with a ratio of about $0.974$ — close to equality but not sharp. The interactive viz below lets you slide $p$ and watch this ratio evolve, including for choices that drive the ratio toward $1$ (the Young’s-inequality saturation case).

Tracing through the proof, equality in Hölder is equivalent to equality in the Young’s-inequality step at $\mu$-almost every point, which (Proposition 1) requires $|\tilde f|^p = |\tilde g|^q$ almost everywhere. Un-normalizing, this becomes: there exist constants $\alpha, \beta \geq 0$, not both zero, such that $\alpha |f|^p = \beta |g|^q$ a.e. Functions of this form — where $|f|^p$ and $|g|^q$ are proportional — are called conjugate pairs.

The sharpness condition is surprisingly fragile: it is not enough for $f$ and $g$ to “look similar.” Consider $f(x) = x^{1/3}$ and $g(x) = x^{1/6}$ on $[0, 1]$ with $p = 3$, $q = 3/2$ (the “power-pair” preset in the explorer above). The powers look naturally matched, but the pointwise check fails: $|f|^p = x$ while $|g|^q = (x^{1/6})^{3/2} = x^{1/4}$, so $|f|^p$ and $|g|^q$ differ by a factor of $x^{3/4}$ — they are not proportional, and the Hölder ratio sits at about $0.9747$ rather than $1$.

To actually get equality at $p = 3$, $q = 3/2$, we need $|f|^p \propto |g|^q$. The simplest way: take $f = x^{\alpha}$ and $g = x^{\beta}$ with $\alpha p = \beta q$, so $3\alpha = (3/2)\beta$, so $\beta = 2\alpha$. Choosing $\alpha = 1/3$ gives $g = x^{2/3}$, and now $|f|^p = x = |g|^q$ exactly. This is the “power-pair-sharp” preset in the explorer — flip to it and the ratio snaps to $1$ to within quadrature precision. Sharp Hölder pairs are the analogue of eigenvalue–eigenvector pairs in a finite-dimensional inner product space: they expose the “tightness directions” of the inequality.

Let $f, g \in L^p(\mu)$ with $1 \leq p \leq \infty$. Then $f + g \in L^p(\mu)$ and

$\|f + g\|_p \;\leq\; \|f\|_p + \|g\|_p.$

The cases $p = 1$ and $p = \infty$ are immediate: $|f + g| \leq |f| + |g|$ pointwise, and integrating (for $p = 1$) or taking the essential supremum (for $p = \infty$) gives the result. We prove the case $1 < p < \infty$ via Hölder.

Step 1: Bound $|f + g|^p$ in terms of $|f|$ and $|g|$. The triangle inequality for absolute values gives $|f + g| \leq |f| + |g|$ pointwise, so

$|f + g|^p \;=\; |f + g| \cdot |f + g|^{p-1} \;\leq\; \bigl( |f| + |g| \bigr) \cdot |f + g|^{p-1}.$

Distributing:

$|f + g|^p \;\leq\; |f| \cdot |f + g|^{p-1} + |g| \cdot |f + g|^{p-1}.$

Step 2: Integrate. Integrating both sides against $\mu$:

$\|f + g\|_p^p \;=\; \int |f + g|^p \, d\mu \;\leq\; \int |f| \cdot |f + g|^{p-1} \, d\mu + \int |g| \cdot |f + g|^{p-1} \, d\mu.$

Step 3: Apply Hölder to each term. Let $q = p / (p - 1)$ be the conjugate exponent of $p$, so $1/p + 1/q = 1$. Apply Hölder’s inequality to the first integral with $f$ in the $L^p$ slot and $|f + g|^{p-1}$ in the $L^q$ slot:

$\int |f| \cdot |f + g|^{p-1} \, d\mu \;\leq\; \|f\|_p \cdot \left( \int |f + g|^{(p-1)q} \, d\mu \right)^{1/q}.$

Compute $(p - 1) q = (p - 1) \cdot p / (p - 1) = p$. So the right side is $\|f\|_p \cdot \|f + g\|_p^{p/q}$. Similarly for the second integral:

$\int |g| \cdot |f + g|^{p-1} \, d\mu \;\leq\; \|g\|_p \cdot \|f + g\|_p^{p/q}.$

Step 4: Combine and divide. Substituting back into Step 2:

$\|f + g\|_p^p \;\leq\; \bigl( \|f\|_p + \|g\|_p \bigr) \cdot \|f + g\|_p^{p/q}.$

If $\|f + g\|_p = 0$ the inequality holds trivially; otherwise we may divide both sides by $\|f + g\|_p^{p/q}$. Note that $p - p/q = p(1 - 1/q) = p \cdot (1/p) = 1$ by conjugacy, so $\|f + g\|_p^p / \|f + g\|_p^{p/q} = \|f + g\|_p^1$. The result is

$\|f + g\|_p \;\leq\; \|f\|_p + \|g\|_p.$

Take $f = \mathbf{1}_{[0, 1/2]}$ and $g = \mathbf{1}_{[1/2, 1]}$ on $[0, 1]$ with Lebesgue measure, and $p = 2$. Compute the three quantities:

$\|f\|_2 \;=\; \left( \int_0^{1/2} 1 \, dx \right)^{1/2} \;=\; \frac{1}{\sqrt{2}},$

$\|g\|_2 \;=\; \left( \int_{1/2}^1 1 \, dx \right)^{1/2} \;=\; \frac{1}{\sqrt{2}},$

$\|f + g\|_2 \;=\; \left( \int_0^1 1 \, dx \right)^{1/2} \;=\; 1.$

(The supports are disjoint, so $f + g$ is the indicator of $[0, 1]$ — except possibly at the single point $x = 1/2$, which has measure zero.) Minkowski says $\|f + g\|_2 \leq \|f\|_2 + \|g\|_2$, which here becomes

$1 \;\leq\; \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \;=\; \sqrt{2} \;\approx\; 1.4142.$

This is a strict inequality with a healthy gap — about $41\%$ slack. The reason: in the $L^2$ norm, “disjoint supports” is not the same as “orthogonal.” The two functions overlap at exactly one point (a measure-zero set), but the $L^2$ geometry treats them as if they were in different “directions” yet still adds them in a non-orthogonal way. The disjoint indicators are orthogonal in the inner-product sense ($\langle f, g \rangle = 0$, by Section 9), and for orthogonal vectors the Pythagorean identity says $\|f + g\|_2^2 = \|f\|_2^2 + \|g\|_2^2$, which is $1 = 1/2 + 1/2$. So the Minkowski inequality holds with strict inequality, but the Pythagorean identity holds with equality — these are two different statements about the same pair, and they are both true.

The Minkowski proof uses the fact that $t \mapsto t^p$ is convex on $[0, \infty)$, which holds for $p \geq 1$. For $0 < p < 1$ the function $t^p$ is concave, and the triangle inequality reverses direction. Concretely, with the same disjoint indicators $f = \mathbf{1}_{[0, 1/2]}$ and $g = \mathbf{1}_{[1/2, 1]}$ but $p = 1/2$:

$\|f\|_{1/2} \;=\; \left( \int_0^{1/2} 1^{1/2} \, dx \right)^{2} \;=\; \left( \frac{1}{2} \right)^2 \;=\; \frac{1}{4},$

$\|g\|_{1/2} \;=\; \frac{1}{4} \quad \text{by symmetry},$

$\|f + g\|_{1/2} \;=\; \left( \int_0^1 1^{1/2} \, dx \right)^2 \;=\; 1^2 \;=\; 1.$

Comparing: $\|f + g\|_{1/2} = 1$ but $\|f\|_{1/2} + \|g\|_{1/2} = 1/2$. So $\|f + g\|_{1/2} > \|f\|_{1/2} + \|g\|_{1/2}$ — the “norm” of the sum is larger than the sum of the “norms.” The triangle inequality fails, $\|\cdot\|_{1/2}$ is not a norm, and $L^{1/2}$ is not a normed space. (It is still a topological vector space; the formula $d(f, g) = \int |f - g|^{1/2} \, d\mu$ defines a translation-invariant metric. But there is no norm.) For this reason the entire $L^p$ theory in the rest of this topic restricts to $p \geq 1$.

For every $1 \leq p \leq \infty$, $\bigl( L^p(\mu), \|\cdot\|_p \bigr)$ is a normed vector space. That is:

$\|f\|_p \;\geq\; 0 \text{ for all } f \in L^p, \text{ with equality iff } f = 0 \text{ in } L^p.$

$\|\alpha f\|_p \;=\; |\alpha| \, \|f\|_p \text{ for all } \alpha \in \mathbb{R} \text{ and } f \in L^p.$

$\|f + g\|_p \;\leq\; \|f\|_p + \|g\|_p \text{ for all } f, g \in L^p.$

Looking back at the construction: the seminorm $\|\cdot\|_p$ on the space of measurable functions has a non-trivial kernel — the set of functions with $\|f\|_p = 0$ — and that kernel is precisely the set of functions that are zero almost everywhere. Quotienting by the kernel collapses the kernel to a single point (the zero element), and on the quotient the seminorm becomes a norm. This is a general construction: in any seminormed vector space, the kernel of the seminorm is a linear subspace, and the quotient by the kernel is automatically a normed space. $L^p$ is one of the most famous applications of this general fact, and it is the reason measure-theoretic functional analysis can use the geometric language of normed vector spaces at all.

A sequence $(f_n)$ in $L^p(\mu)$ is Cauchy if for every $\varepsilon > 0$ there exists $N$ such that $\|f_n - f_m\|_p < \varepsilon$ for all $n, m \geq N$.

Equivalently, $\|f_n - f_m\|_p \to 0$ as $\min(n, m) \to \infty$.

For every $1 \leq p \leq \infty$ and every measure space $(\Omega, \mathcal{F}, \mu)$, the space $L^p(\mu)$ is complete: every Cauchy sequence in $L^p(\mu)$ converges to an element of $L^p(\mu)$.

In particular, $L^p(\mu)$ with the norm $\|\cdot\|_p$ is a Banach space.

We prove the case $1 \leq p < \infty$. The case $p = \infty$ is a separate (easier) argument that we sketch in Remark 7. Let $(f_n)$ be a Cauchy sequence in $L^p$. We will construct a candidate limit, show that a subsequence converges to it, and then use the Cauchy property to upgrade subsequence convergence to full-sequence convergence.

Step 1: Extract a fast-converging subsequence. Since $(f_n)$ is Cauchy, we can choose indices $n_1 < n_2 < n_3 < \cdots$ such that

$\|f_{n_{k+1}} - f_{n_k}\|_p \;<\; 2^{-k} \quad \text{for every } k \geq 1.$

This is possible because the Cauchy condition with $\varepsilon_k = 2^{-k}$ gives an index $N_k$ such that $\|f_n - f_m\|_p < 2^{-k}$ for $n, m \geq N_k$; we just choose $n_{k+1} > \max(n_k, N_k)$.

Step 2: Bound the partial-sum-of-differences. Define the non-negative function

$g_N(x) \;=\; \sum_{k=1}^N |f_{n_{k+1}}(x) - f_{n_k}(x)|.$

Each $g_N$ is measurable and non-negative. By Minkowski’s inequality applied $N$ times,

$\|g_N\|_p \;\leq\; \sum_{k=1}^N \|f_{n_{k+1}} - f_{n_k}\|_p \;<\; \sum_{k=1}^N 2^{-k} \;<\; 1.$

So the $L^p$ norms of the partial sums $g_N$ are uniformly bounded by $1$.

Step 3: Pass to the monotone limit. The sequence $(g_N)$ is monotone increasing — adding another non-negative term cannot decrease the sum. By the Monotone Convergence Theorem (Topic 26, Section 5) applied to $(g_N^p)$, the pointwise limit

$g(x) \;=\; \lim_{N \to \infty} g_N(x) \;=\; \sum_{k=1}^\infty |f_{n_{k+1}}(x) - f_{n_k}(x)|$

exists in $[0, \infty]$ at every $x$ and satisfies

$\int g^p \, d\mu \;=\; \lim_{N \to \infty} \int g_N^p \, d\mu \;=\; \lim_{N \to \infty} \|g_N\|_p^p \;\leq\; 1^p \;=\; 1.$

So $g \in L^p$, and in particular $g(x) < \infty$ for $\mu$-almost every $x$. This is the key consequence of the bound: the infinite sum of absolute differences is finite a.e.

Step 4: Construct the candidate limit $f$. Where $g(x) < \infty$ — i.e., on a set of full measure — the series

$f_{n_1}(x) + \sum_{k=1}^\infty \bigl( f_{n_{k+1}}(x) - f_{n_k}(x) \bigr)$

is absolutely convergent. (This is the standard “absolute convergence implies convergence” fact for real-valued series, applied at each $x$.) The partial sums of this series are exactly $f_{n_{N+1}}(x)$ — the series telescopes — so we define

$f(x) \;=\; \lim_{k \to \infty} f_{n_k}(x) \quad \text{for $\mu$-a.e. } x,$

and $f(x) = 0$ on the null set where the series fails to converge. The function $f$ is measurable (as the a.e.-limit of measurable functions, redefined on a null set).

Step 5: $f \in L^p$. We need to check that the candidate limit actually lives in $L^p$. The triangle inequality gives, for every $k \geq 1$,

$|f_{n_k}(x) - f_{n_1}(x)| \;\leq\; \sum_{j=1}^{k-1} |f_{n_{j+1}}(x) - f_{n_j}(x)| \;\leq\; g(x).$