Home > Analysis, Teaching > A nice application of Fatou’s Lemma

A nice application of Fatou’s Lemma

Let me show you an exciting technique to prove some convergence statements using exclusively functional inequalities and Fatou’s Lemma. The following are two classic problems solved this way. Enjoy!

Exercise 1 Let {(X, \mathcal{F}, \mu)} be a measurable space and suppose {\{f_n\}_{n\in \mathbb{N}}} is a sequence of measurable functions in {L_1(\mu)} that converge almost everywhere to a function {f \in L_1(\mu)}, and such that the sequence of norms {\big\{ \lVert f_n \rVert_1 \big\}_{n \in \mathbb{N}}} converges to {\lVert f \rVert_1}. Prove that the sequence of integrals {\int_E \lvert f_n \rvert\, d\mu} converges to the integral {\int_E \lvert f \rvert\, d\mu} for every measurable set {E}.

Proof: Note first that

\displaystyle  \begin{array}{rcl}  \lvert f_n - f\rvert\, \boldsymbol{1}_E \leq \lvert f_n - f \rvert \leq \lvert f_n \rvert + \lvert f \rvert. \end{array}

Set then {g_n = \lvert f_n \rvert + \lvert f \rvert - \lvert f_n-f \rvert\, \boldsymbol{1}_E} (which are non-negative functions) and apply Fatou’s Lemma to that sequence. We have then

\displaystyle  \begin{array}{c}  \int_X \liminf_n g_n\, d\mu \leq \liminf_n \int_X g_n\, d\mu \\ \\ \int_X \liminf_n \big( \lvert f_n \rvert + \lvert f \rvert - \lvert f_n - f \rvert\, \boldsymbol{1}_E \big)\, d\mu \leq \liminf_n \int_X \big( \lvert f_n \rvert + \lvert f \rvert - \lvert f_n - f \rvert\, \boldsymbol{1}_E \big)\, d\mu \\ \\ \int_X \big( 2\lvert f \rvert - \limsup_n \lvert f_n-f \rvert\, \boldsymbol{1}_E\big)\, d\mu \leq \liminf_n \int_X \lvert f_n \rvert\, d\mu + \lVert f \rVert_1 - \limsup_n \int_E \lvert f_n -f \rvert\, d\mu \\ \\ 2\lVert f\rVert_1 \leq 2\lVert f \rVert_1 - \limsup_n \int_E \lvert f_n - f\rvert\, d\mu, \end{array}

which implies

\displaystyle  \begin{array}{rcl}  0 \leq \liminf_n \int_E \lvert f_n-f \rvert\, d\mu \leq \limsup_n \int_E \lvert f_n-f \rvert\, d\mu \leq 0. \end{array}

It must then be {\int_E \lvert f_n - f \rvert\, d\mu = 0}. But this proves the statement, since

\displaystyle  \begin{array}{c}  \bigg\lvert \int_E \lvert f_n\rvert\, d\mu - \int_E \lvert f \rvert\, d\mu \bigg\rvert = \bigg\lvert \int_E \big( \lvert f_n \rvert - \lvert f \rvert \big)\, d\mu \bigg\rvert \\  \leq \int_E \Big\lvert \lvert f_n \rvert - \lvert f \rvert \Big\rvert\, d\mu \leq \int_E \lvert f_n - f \rvert\, d\mu \end{array}

\Box

Exercise 2 Let {(X, \mathcal{F}, \mu)} be a finite measure space and let {1<p<\infty}. Suppose that {\{ f_n \}_{n \in \mathbb{N}}} is a sequence of measurable functions in {L_p(\mu)} whose norms are uniformly bounded in {n} and which converge almost everywhere to a function {f}. Prove that the sequence {\big\{ \int_X f_ng\, d\mu \big\}_{n \in \mathbb{N}}} converges to {\int_x fg\, d\mu} for all {g \in L_q(\mu)} where {q} is the conjugate exponent of {p}.

Proof: The proof is very similar to the previous problem. We start by noticing that under the hypotheses of the problem,

\displaystyle  \begin{array}{c}  \bigg\lvert \int_x f_ng\, d\mu - \int_X fg\, d\mu \bigg\rvert = \bigg\lvert \int_X (f_n -f)g\, d\mu \bigg\rvert \\ \leq \lVert (f_n-f)g \rVert_1 \leq \lVert f_n - f \rVert_p\, \lVert g \rVert_q. \end{array}

If we prove that {\lim_n \lVert f_n-f \rVert_p = 0}, we are done.

We will achieve this by using the convexity of {\lvert \cdot \rvert^p}, since in that case it is

\displaystyle  \begin{array}{rcl}  \frac{\lvert f_n - f \rvert^p}{2^p} \leq \tfrac{1}{2} \lvert f_n \rvert^p + \tfrac{1}{2} \lvert f \rvert^p. \end{array}

hence,

\displaystyle  \begin{array}{rcl}  \lvert f_n -f \rvert^p \leq 2^{p-1} \big( \lvert f_n \rvert^p + \lvert f \rvert^p \big). \end{array}

Set then {g_n = 2^{p-1} \big( \lvert f_n\rvert^p + \lvert f \rvert^p\big) - \lvert f_n-f \rvert^p} (which are non-negative functions) and apply Fatou’s Lemma as before. \Box

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