Convergence in Lp Norm

Author

John Robin Inston

Published

September 25, 2026

1 Convergence in Norm

Recall that a norm is a function \(\Vert \cdot\Vert\) from a vector (function) space \(V\) to \(\mathbb{R}\) satisfying, for all \(f,g\in V\) and \(a \in \mathbb{R}\):

  1. \(\Vert f\Vert \geq 0\), and \(\Vert f\Vert=0\) if and only if \(f=0\);
  2. \(\Vert af\Vert = |a|\Vert f\Vert\);
  3. \(\Vert f+g \Vert \leq\Vert f\Vert +\Vert g\Vert\) (triangle inequality).

A sequence \(\{f_n\}_{n\geq 1}\) in a normed space \((V, \Vert\cdot\Vert)\) converges in norm to \(f\), written \(f_n\stackrel{\Vert\cdot\Vert}{\to}f\), if \[ \lim_{n\to\infty}\Vert f_n-f\Vert=0. \]

2 Convergence in \(L^p\) Norm (r-th Mean)

The \(L^p\) norm is the norm linked to probability spaces and expectations, see Lp Spaces and Lp Norm. For a random variable \(X\) on \((\Omega,\mathcal{F},\mathbb{P})\) and \(p\geq1\), \[ \Vert X\Vert_{p}=[\mathbb{E}|X|^p]^{1/p}=\left(\int _{\Omega}|X|^p \, d\mathbb{P}\right)^{1/p} . \]

A sequence of random variables \(\{ X_{n} \}_{n\in\mathbb{N}}\) with \(\mathbb{E}|X_n|^p<\infty\) converges in \(L^p\) norm, or in \(p\)-th mean, to \(X\), written \(X_{n}\stackrel{L^p}{\to}X\), if \[ \lim_{ n \to \infty }\Vert X_{n}-X\Vert_{p}=0,\quad\text{equivalently}\quad\lim_{ n \to \infty } \mathbb{E}|X_{n}-X|^p = 0. \]

Convergence in \(p\)-th mean is convergence of integrals (expectations) and is exactly convergence in norm on \(L^p\). Random variables can also converge almost surely, in probability and in law; the relationships between all modes are in Convergence of Random Variables.

3 Relationships

If \(X_{n}\stackrel{L^s}{\to}X\) and \(1\leq r<s<\infty\) then \(X_{n}\stackrel{L^r}{\to}X\). In other words, higher orders of convergence imply lower orders.

Proof. By Lyapunov Inequality applied to \(|X_n-X|\), \[ (\mathbb{E}|X_n-X|^r)^{1/r}\leq(\mathbb{E}|X_n-X|^s)^{1/s}\to0. \] Alternatively, apply Jensen Inequality with the convex function \(f(x)=x^{s/r}\) to \(|X_n-X|^r\).

The converse does not hold in general.

If \(X_{n}\stackrel{L^1}{\to}X\) then \(X_{n}\stackrel{\mathbb{P}}{\to}X\).

Proof. By the Markov Inequality, for every \(\epsilon>0\) \[ \mathbb{P}(|X_{n}-X|>\epsilon)\leq \frac{1}{\epsilon}\cdot \mathbb{E}|X_{n}-X|\to0. \]

On \([0,1]\) with Lebesgue measure let \(X_n=n\mathbb{1}_{(0,1/n)}\). Then \(X_n\to0\) almost surely and in probability, but \(\mathbb{E}|X_n|=1\) for all \(n\), so \(X_n\not\to0\) in \(L^1\).

The converse holds under Uniform Integrability.

For \(p>0\) assume that \(X_n \in L^p\), \(n \geq 1\) and \(X_{n}\stackrel{\mathbb{P}}{\to} X\). The following are equivalent:

  1. \(\{ |X_{n}|^p,~n\geq 1 \}\) is uniformly integrable;
  2. \(X_{n}\stackrel{L^p}{\to}X\) and \(X\in L^p\); and
  3. \(\lim_{n \to \infty}\mathbb{E}[|X_{n}|^p]=\mathbb{E}[|X|^p]<\infty\).

4 References

  • A. Gut, Probability: A Graduate Course, Springer.
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