Ngô Quốc Anh

April 17, 2010

A Theorem of Banach and Saks

Filed under: Giải tích 8 (MA5206) — Ngô Quốc Anh @ 23:39

According to Banach and Saks, every bounded sequence in L^p or \ell^p (1<p< \infty) has a subsequence whose Cesaro-means converge strongly. More generally, every uniformly convex Banach space possesses this so-called Banach-Saks property, as shown by Kakutani. In particular, every Hilbert space has this property. In nonlinear analysis, by utilizing a duality mapping some assertions which are valid in the case of Hilbert spaces are extended to the case of special classes of Banach spaces. Especially in the case of Banach spaces with a uniformly convex conjugate space, such extentions are often obtained since a duality mapping is uniformly strongly continuous on each bounded subset of such a Banach space.

The Banach-Saks theorem in L^2 states that

Theorem (Banach-Saks for L^2 spaces). Given in L^2 a sequence \{f_n\}_n which converges weakly to an element f, we can select a subsequence \{f_{n_k}\}_k such that the arithmetic means

\displaystyle\frac{{{f_{{n_1}}} + {f_{{n_2}}} + \cdots + {f_{{n_k}}}}}{k}

converge in strongly to f.

This theorem is due to the two Polish geometers S.  Banach and S. Saks, whose work and, in particular, the importance of whose research in the topics treated in this book are widely acknowledged.

Proof. Replacing f_n by f_n -f, we can assume f = 0. We shall successively choose the n_k in the following manner. Beginning for definiteness with n_1 = 1, let n_2 be an index or, if we wish, the first index such that

\displaystyle |(f_1,f_n)| \leqslant 1;

this choice is possible since (f_1, f_n) \to 0 as n\to \infty. In general, after having chosen f_{n_1}, f_{n_2},...,f_{n_k} we choose n_{k+1} > n_k so that

\displaystyle\left| {\left( {{f_{{n_1}}},{f_{{n_{k + 1}}}}} \right)} \right| \leqslant \frac{1}{k},...,\left| {\left( {{f_{{n_n}}},{f_{{n_{k + 1}}}}} \right)} \right| \leqslant \frac{1}{k},

which is possible since (f_{n_i}, f_n) \to 0 as i=\overline{1,k} and n\to \infty. Since, furthermore, the norms \|f_n\| form a bounded sequence, say \|f_n\| \leqslant B, it follows by expanding the inner product that

\displaystyle {\left\| {\frac{{{f_{{n_1}}} + {f_{{n_2}}} + \cdots + {f_{{n_k}}}}}{k}} \right\|^2} \leqslant \frac{{k{B^2} + 2 \times 1 + 4 \times \frac{1}{2} + \cdots + 2(k - 1) \times \frac{1}{{k - 1}}}}{{{k^2}}} < \frac{{{B^2} + 2}}{k}

which implies

\displaystyle {\left\| {\frac{{{f_{{n_1}}} + {f_{{n_2}}} + \cdots + {f_{{n_k}}}}}{k}} \right\|^2} \to 0

as k\to \infty.

Corollary. If a linear set, or more generally a convex set, in L^2 is closed in the sense of weak convergence, it is also closed in the sense of strong convergence.

In a Hilbert space, we still have the following version:

Theorem (Banach-Saks for Hilbert spaces). Every bounded sequence \{x_n\}_n contains a subsequence \{x_{n_k}\}_k and a point x such that

\displaystyle\frac{{{x_{{n_1}}} + {x_{{n_2}}} + \cdots +  {x_{{n_k}}}}}{k}

converges strongly to x as k goes to infinity.

In a Banach space, we have the following version:

Theorem (Banach-Saks for Banach spaces). If a sequence of elements \{x_n\}_n converges weakly to an element x, then there exists some subsequence of elements \{x_{n_k}\}_k such that the averages

\displaystyle\frac{{{x_{{n_1}}} + {x_{{n_2}}} + \cdots +   {x_{{n_k}}}}}{k}

converges strongly to x as k goes to infinity.

It is natural enough to ask whether the same property is true for functionals that converge in the weak-* topology, but unfortunately (as is well known) the answer is no. Consider the sequence of functions

\displaystyle e^{i\theta}, e^{2\theta},...

in L^1[0, 2\pi) to be a sequence of linear functionals on L^1[0, 2\pi). This sequence converges in the weak-* topology to 0 by the Riemann-Lebesgue lemma, yet for any subsequence, the average will have a norm of 1 since each function approaches 1 as \theta tends to 0.

Cf: S. Banach and S. Saks: Sur la convergence forte dans les champs L^p, Studia Math. 2 (1930), pp. 51-57.

Leave a Comment »

No comments yet.

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Create a free website or blog at WordPress.com.

%d bloggers like this: