Lp spaces §

Why equivalence classes? Two functions that differ only on a set of measure zero have the same integral and the same norm. Working with equivalence classes makes the norm well-defined (a true norm, not just a seminorm).

The norm. For $1 \leq p < \infty$, $\|\cdot\|_p$ satisfies:

1. $\|f\|_p \geq 0$, with $\|f\|_p = 0 \iff f = 0$ a.e.
2. $\|\alpha f\|_p = |\alpha| \, \|f\|_p$
3. Minkowski inequality (triangle inequality): $\|f + g\|_p \leq \|f\|_p + \|g\|_p$

Completeness. $L^p$ is a Banach space for all $1 \leq p \leq \infty$: every Cauchy sequence in $\|\cdot\|_p$ converges to an element of $L^p$.

Separability. If $\mu$ is $\sigma$-finite, then $L^p$ is separable for $1 \leq p < \infty$ (simple functions with rational coefficients on a countable algebra generate a dense subset). In contrast, $L^\infty$ is never separable on a non-atomic space.

Uniform convexity. $L^p$ is uniformly convex if and only if $1 < p < \infty$. This geometric property underpins the norm-convergence characterization in the Convergence section and implies that $L^p$ is reflexive for $1 < p < \infty$.

Unit balls in two dimensions (the same picture governs $L^p$ on a two-atom measure space). For $1 < p < \infty$ the ball is strictly convex: no flat spots, so midpoints of distinct boundary points lie strictly inside — this is uniform convexity. At $p = 1$ and $p = \infty$ the ball has flat sides, and uniform convexity fails.

Special case $p = 2$: $L^2$ is a Hilbert space with inner product $\langle f, g \rangle = \int f \bar{g} \, d\mu$.

Proof sketch. For $p, q > 1$, apply Young’s inequality $ab \leq \frac{a^p}{p} + \frac{b^q}{q}$ to $a = |f|/\|f\|_p$, $b = |g|/\|g\|_q$, then integrate.

Consequences:

• $L^p \cdot L^q \subseteq L^1$ (product of $L^p$ and $L^q$ functions is integrable)
• $\|f\|_p \leq \mu(X)^{1/p - 1/q} \, \|f\|_q$ for $p < q$ (on finite measure spaces, $L^q \subseteq L^p$ — see Nesting of $L^p$ spaces)

Example. $p = 2$, $q = 2$: Cauchy–Schwarz inequality $\left|\int fg \, d\mu\right| \leq \|f\|_2 \, \|g\|_2$.

For $p = 1$, this is the triangle inequality for integrals: $\int |f+g| \leq \int |f| + \int |g|$.

For $p > 1$, the proof uses Hölder’s inequality. Write $|f+g|^p = |f+g|^{p-1} \cdot |f+g|$, split into $|f+g|^{p-1}|f|$ and $|f+g|^{p-1}|g|$, apply Hölder with exponents $p$ and $q = p/(p-1)$ to each term, and factor.

This is what makes $\|\cdot\|_p$ a true norm (rather than just a seminorm) and $L^p$ a normed vector space.

Concrete example. Let $f = \mathbf{1}_{[0,1]}$ and $g = \mathbf{1}_{[1,2]}$ on $\mathbb{R}$ with Lebesgue measure, and take $p = 2$. Then:

So $\|f+g\|_2 = \sqrt{2} < 2 = \|f\|_2 + \|g\|_2$. The inequality is strict because $f$ and $g$ have disjoint supports (they are “orthogonal” in the $L^2$ sense). Equality in Minkowski holds if and only if $f$ and $g$ are non-negative scalar multiples of each other a.e.

Proof of the inclusion. For $q < \infty$, apply Hölder’s inequality with exponents $r = q/p > 1$ and its conjugate $r' = q/(q-p)$ to the functions $|f|^p$ and $\mathbf{1}_X$:

Taking the $1/p$-th root gives $\|f\|_p \leq \mu(X)^{1/p - 1/q} \|f\|_q$. For $q = \infty$ the inequality $\|f\|_p \leq \mu(X)^{1/p} \|f\|_\infty$ is immediate from $|f| \leq \|f\|_\infty$ a.e.

Intuition. On a finite measure space, a spike of height $H$ on a tiny set contributes little to $\int |f|^p$ when $p$ is small, but dominates when $p$ is large. So requiring $\int |f|^q < \infty$ for a large $q$ is a strong condition: it forbids tall spikes. Requiring $\int |f|^p < \infty$ for a small $p$ is weaker. Hence $L^q \subseteq L^p$ for $p < q$. The inclusion is strict in general: $f(x) = x^{-1/2}$ on $(0,1)$ belongs to $L^1$ (integral $2$) but not to $L^2$ ($\int_0^1 x^{-1} dx = \infty$).

Example. On $(0,1)$, take $f(x) = x^{-1/3}$ (the function of the Quiz below). Here $f \notin L^p$ for $p \geq 3$ and $\|f\|_\infty = \infty$: the norms blow up exactly because the spike at $0$ is unbounded, consistent with the limit theorem. For a bounded illustration, let $g = \min(x^{-1/4}, 10)$ on $(0,1)$, so $\|g\|_\infty = 10$; then $\|g\|_p$ increases toward $10$ as $p \to \infty$, since the values near the cap $10$ weigh ever more heavily in $\int |g|^p$.

What survives on arbitrary measure spaces is the interpolation inequality: if $f \in L^p \cap L^q$ with $p < r < q$, then $f \in L^r$ and $\|f\|_r \leq \|f\|_p^{\theta} \|f\|_q^{1-\theta}$ where $\frac{1}{r} = \frac{\theta}{p} + \frac{1-\theta}{q}$.

Proof sketch of the limit $\|f\|_p \to \|f\|_\infty$. Assume $0 < \|f\|_\infty < \infty$ and $\mu(X) < \infty$. Upper bound: $\|f\|_p \leq \mu(X)^{1/p} \|f\|_\infty \to \|f\|_\infty$. Lower bound: for every $\varepsilon > 0$ the set $A_\varepsilon = \{|f| > \|f\|_\infty - \varepsilon\}$ has positive measure, so $\|f\|_p \geq (\|f\|_\infty - \varepsilon)\, \mu(A_\varepsilon)^{1/p}$, and letting $p \to \infty$ then $\varepsilon \to 0$ gives $\liminf_p \|f\|_p \geq \|f\|_\infty$. (On an infinite measure space the limit can still hold, but the upper bound needs an extra argument; the statement above is the standard finite-measure form.)

Example. On $(0,1)$, take $f(x) = x^{-1/3}$ (the function of the Quiz below). Here $f \notin L^p$ for $p \geq 3$, and indeed $\|f\|_\infty = \infty$: the norm “blows up” exactly because the spike at $0$ is unbounded, consistent with the limit theorem. For a bounded example, $f(x) = x^{-1/4}$ has $\|f\|_p = \left(\frac{4}{4-p}\right)^{1/p}$ for $p < 4$, which grows to $\|f\|_\infty = \infty$… take instead the bounded $f(x) = x^{-1/4}$ truncated: $g = \min(x^{-1/4}, 10)$ on $(0,1)$ has $\|g\|_\infty = 10$ and $\|g\|_p \uparrow 10$ as $p \to \infty$.

Why the naive “boundedness” condition fails. The statement “$f_n \to f$ in $L^p$ iff $f_n \to f$ in measure and $\|f_n\|_p$ is bounded” is false. Counterexample: $f_n = n^{1/p} \cdot \mathbf{1}_{(0,1/n)}$ on $(0,1)$. Then $f_n \to 0$ in measure (the support shrinks to a point) and $\|f_n\|_p = 1$ for all $n$ (bounded), yet $\|f_n - 0\|_p = 1 \not\to 0$. The mass concentrates near 0 without the norm vanishing. What is needed is not mere boundedness but uniform integrability of $\{|f_n|^p\}$.

The easy direction, and what $L^p$ convergence always implies. If $\|f_n - f\|_p \to 0$, then by the Minkowski inequality $|\|f_n\|_p - \|f\|_p| \leq \|f_n - f\|_p \to 0$, so the norms converge; and Chebyshev’s inequality gives $\mu\{|f_n - f| > \varepsilon\} \leq \varepsilon^{-p} \|f_n - f\|_p^p \to 0$, so $f_n \to f$ in measure (hence a subsequence converges a.e.). These one-direction facts hold for all $1 \leq p < \infty$ and are the “$\Rightarrow$” half of both theorems above.

Why the general theorem needs no norm condition. In the general theorem, conditions (1) and (2) already force $\|f_n\|_p^p \to \|f\|_p^p$: apply the Vitali convergence theorem to the functions $|f_n|^p$, which converge in measure to $|f|^p$ and are uniformly integrable by (2). Hence $\limsup_n \|f_n\|_p^p \leq \|f\|_p^p$ (in fact equality) comes for free, and listing it as a third condition would be redundant.

Relationships between modes of convergence (on finite measure spaces):

The reverse implications do not hold in general, but:

• DCT (see Lebesgue integral) gives: a.e. convergence + domination $\implies L^p$ convergence (for $p < \infty$ on finite measure spaces)
• Uniform integrability + convergence in measure $\implies L^1$ convergence (Vitali convergence theorem)

Example. $f_n = n \cdot \mathbf{1}_{(0, 1/n)}$ on $(0,1)$. Then $f_n \to 0$ a.e., but $\|f_n\|_1 = 1$ for all $n$, so $f_n \not\to 0$ in $L^1$. The mass “concentrates” near 0.

In $L^\infty$: $\|f_n - f\|_\infty \to 0$ means uniform convergence a.e. (essential uniform convergence). This is strictly stronger than $L^p$ convergence for finite $p$.

Why $L^1$ and $L^\infty$ fail reflexivity. $L^1$ is not reflexive because its dual $(L^1)^* = L^\infty$ has a larger dual than $L^1$ itself. $L^\infty$ is not reflexive because it contains an isometric copy of $\ell^\infty$ (on a non-atomic space), which is not reflexive.

Connection to Hölder. The duality pairing $\langle f, g \rangle = \int fg \, d\mu$ is precisely the one bounded by Hölder’s inequality. The isometric isomorphism means the operator norm of the functional equals the $L^q$ norm of the representing function.

Practical consequence. In optimization and PDE theory, weak convergence in $L^p$ (i.e., convergence against all test functions in $L^q$) is the natural compactness tool: bounded sequences in $L^p$ for $1 < p < \infty$ have weakly convergent subsequences (Banach–Alaoglu + reflexivity).

Convergence of moments. If $X_n \to X$ in $L^p$, then $\mathbb{E}[|X_n|^p] \to \mathbb{E}[|X|^p]$ (the norm is continuous), and more generally $\mathbb{E}[|X_n|^r] \to \mathbb{E}[|X|^r]$ for every $0 < r \leq p$. This is why $L^p$ convergence is the standard hypothesis behind “convergence of moments” results: it rules out the mass-concentrating pathologies (spikes on vanishing sets) that make a.e. convergence alone too weak.

The $L^2$ inner product gives the covariance. For centered random variables ($\mathbb{E}[X] = \mathbb{E}[Y] = 0$), the $L^2$ inner product $\langle X, Y \rangle = \mathbb{E}[XY]$ is exactly the covariance $\mathrm{Cov}(X, Y)$, and $\|X\|_2^2 = \mathrm{Var}(X)$. The Cauchy–Schwarz inequality becomes $|\mathrm{Cov}(X,Y)| \leq \sqrt{\mathrm{Var}(X)\,\mathrm{Var}(Y)}$, which is the source of the bound $|\rho| \leq 1$ on the correlation coefficient. Orthogonality in $L^2$ is uncorrelatedness — the geometric foundation of least squares, the law of total variance, and the projection arguments behind conditional expectation (see Random variables and expectation).

Why analysts care. Martingale convergence, the law of large numbers in its strong form, and the basic limit theorems of statistics are all most cleanly stated as $L^p$ statements with a separate a.e.-convergence clause — the two modes of convergence are genuinely independent, and $L^p$ is the one that carries quantitative control.

Step 1: Compute $\|f\|_p$.

This integral converges if and only if $-p/3 > -1$, i.e., $p < 3$.

For $p < 3$:

So $\|f\|_p = \left(\frac{3}{3-p}\right)^{1/p}$.

Step 2: Check $p = \infty$.

$\|f\|_\infty = \text{ess sup}_{x \in (0,1)} x^{-1/3} = \infty$ (the function is unbounded near 0).

Conclusion. $f \in L^p(0,1)$ if and only if $1 \leq p < 3$. In particular, $f \in L^1$ and $f \in L^2$, but $f \notin L^3$ and $f \notin L^\infty$.