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Convergence in Mean — Definition, Formula & Examples

Convergence in mean is a way of saying that a sequence of functions gets closer to a limiting function when you measure the distance using an integral of their difference raised to a power. Specifically, the integrated average of the difference shrinks to zero.

A sequence of functions {fn}\{f_n\} converges in the pp-th mean (or in LpL^p) to a function ff on a measure space (X,μ)(X, \mu) if lim⁡n→∞∫X∣fn−f∣p dμ=0\lim_{n \to \infty} \int_X |f_n - f|^p \, d\mu = 0, where p≥1p \geq 1. When p=2p = 2, this is called convergence in mean square; when p=1p = 1, it is simply called convergence in mean.

Key Formula

lim⁡n→∞∫X∣fn(x)−f(x)∣p dμ=0\lim_{n \to \infty} \int_X |f_n(x) - f(x)|^p \, d\mu = 0
Where:
  • fnf_n = The $n$-th function in the sequence
  • ff = The limiting function
  • pp = A real number $\geq 1$ specifying which $L^p$ space is used
  • μ\mu = The measure on the space $X$ (e.g., Lebesgue measure)
  • XX = The domain of integration

How It Works

To check convergence in mean, you compute the LpL^p norm of the difference fn−ff_n - f and verify it tends to zero. In practice, you evaluate ∥fn−f∥p=(∫∣fn−f∣p dμ)1/p\| f_n - f \|_p = \left( \int |f_n - f|^p \, d\mu \right)^{1/p} for each nn and take the limit. This type of convergence is stronger than convergence in probability but weaker than uniform convergence. It is widely used in Fourier analysis, where partial sums of a Fourier series converge in L2L^2 mean to the original function, even when pointwise convergence fails.

Worked Example

Problem: Let fn(x)=xnf_n(x) = x^n on the interval [0,1][0,1] with Lebesgue measure. Determine whether fnf_n converges in mean (i.e., in L1L^1) to f(x)=0f(x) = 0.
Step 1: Write the L1L^1 norm of the difference fn−ff_n - f.
∫01∣fn(x)−0∣1 dx=∫01xn dx\int_0^1 |f_n(x) - 0|^1 \, dx = \int_0^1 x^n \, dx
Step 2: Evaluate the integral.
∫01xn dx=xn+1n+1∣01=1n+1\int_0^1 x^n \, dx = \frac{x^{n+1}}{n+1}\Big|_0^1 = \frac{1}{n+1}
Step 3: Take the limit as n→∞n \to \infty.
lim⁡n→∞1n+1=0\lim_{n \to \infty} \frac{1}{n+1} = 0
Answer: Since the integral tends to 00, the sequence fn(x)=xnf_n(x) = x^n converges in mean (L1L^1) to f(x)=0f(x) = 0 on [0,1][0,1].

Another Example

Problem: Let fn(x)=n⋅1[0,1/n2](x)f_n(x) = n \cdot \mathbf{1}_{[0, 1/n^2]}(x) on [0,1][0,1]. Does fnf_n converge in mean square (L2L^2) to f=0f = 0?
Step 1: Compute the L2L^2 norm squared of the difference.
∫01∣fn(x)∣2 dx=∫01/n2n2 dx=n2⋅1n2=1\int_0^1 |f_n(x)|^2 \, dx = \int_0^{1/n^2} n^2 \, dx = n^2 \cdot \frac{1}{n^2} = 1
Step 2: Check the limit.
lim⁡n→∞1=1≠0\lim_{n \to \infty} 1 = 1 \neq 0
Answer: The L2L^2 norm does not tend to 00, so fnf_n does not converge in mean square to 00, even though fn(x)→0f_n(x) \to 0 pointwise for every x∈(0,1]x \in (0,1].

Visualization

Why It Matters

Convergence in mean is central to courses in real analysis, functional analysis, and probability theory. In signal processing, Fourier series of square-integrable signals converge in L2L^2 mean, guaranteeing that approximations capture nearly all the signal's energy. Probability relies on LpL^p convergence to establish properties of estimators and to prove limit theorems.

Common Mistakes

Mistake: Assuming pointwise convergence guarantees convergence in mean.
Correction: Pointwise convergence alone does not ensure the integral of ∣fn−f∣p|f_n - f|^p goes to zero. You must verify the integral condition directly, or use a theorem like the Dominated Convergence Theorem.
Mistake: Confusing convergence in L1L^1 with convergence in L2L^2 (mean square).
Correction: These are different conditions. On a finite measure space, L2L^2 convergence implies L1L^1 convergence (by the Cauchy–Schwarz inequality), but the reverse is false. Always specify the value of pp.

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