Lp spaces · Definition

Riesz–Fischer theorem

measure theory and integrationlp spaces

Before you start you need to know

The \( L^p \) spaces are the natural setting of modern functional analysis. This theorem guarantees that Cauchy sequences do not escape from them.

Definition

The Riesz-Fischer theorem states that the space \( L^p(\mu) \) is complete for \( 1 \le p \le \infty \), that is, every Cauchy sequence in it converges to an element of \( L^p(\mu) \).

\[\lim_{n,m\to\infty}\|f_n-f_m\|_p=0 \implies \exists f\in L^p:\ \|f_n-f\|_p\to 0\]

If \( (f_n) \) is Cauchy in \( L^p(\mu) \), one extracts a subsequence \( (f_{n_k}) \) with \( \|f_{n_{k+1}}-f_{n_k}\|_p < 2^{-k} \). By Minkowski's inequality, the function \( g=\sum_k |f_{n_{k+1}}-f_{n_k}| \) satisfies \( \|g\|_p \le 1 \), so it is finite almost everywhere. The series \( f_{n_1}+\sum_k (f_{n_{k+1}}-f_{n_k}) \) converges absolutely a.e. to a limit \( f \).

Take \( \mu \) to be Lebesgue measure on \([0,1]\) and \( f_n = n\,\mathbf{1}_{(0,1/n)} \) in \( L^1 \). Since \( \|f_n-f_m\|_1 \) does not tend to zero, this sequence is not Cauchy; instead \( f_n=\mathbf{1}_{(0,1/n)} \) in \( L^1 \) is Cauchy and converges to \( 0 \) in norm. Fatou's lemma then verifies that \( \|f_n-f\|_p \to 0 \).

A typical error is to conclude that pointwise convergence of the subsequence implies convergence in \( L^p \) without proving the bound on \( g \). Without Minkowski's inequality there is no control over \( g \), and the argument collapses.

Think about it

Why is \( C[0,1] \) with the \( L^1 \) norm not a complete space?

Because a sequence of continuous functions can converge in \( L^1 \) to a discontinuous function, such as an indicator, which is not in \( C[0,1] \).

Exercises 10

References

  • Gerald B. Folland, Real Analysis: Modern Techniques and Their Applications, 2nd edition, Wiley, 1999.
  • H. L. Royden, P. M. Fitzpatrick, Real Analysis, 4th edition, Pearson, 2010.