Mazur's Theorem
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Mazur's Theorem

Theorem (Mazur's Theorem): Let $X$ be a normed linear space and let $K$ be a convex subset of $X$. Then $K$ is norm closed if and only if $K$ is weakly closed.

Mazur's Theorem says that the concept of norm closed and weakly closed is the same for convex subsets of a normed linear space $X$. Since we know that the weak and norm topologies on a normed linear space $X$ coincide if and only if $X$ if finite-dimensional, we see that Mazur's theorem is only really interesting when discussing infinite-dimensional normed linear spaces.

  • Proof: $\Rightarrow$ Clearly the norm-closed $\emptyset$ is also weakly closed. So let $K$ be a nonempty, norm-closed convex subset of $X$. Let $x_0 \in X \setminus K$. Since $X$ is a normed linear space by one of the corollaries on The Separation Theorems page, there exists a bounded linear functional $f \in X^*$ such that:
(1)
\begin{align} f(x_0) < c = \inf_{k \in K} f(k) \end{align}
  • Consider the set $N = \{ x \in X : f(x) < c \}$. Note that $x_0 \in N$. Furthermore, $N \subset X \setminus K$ and $N$ is weakly open, since:
  • Hence $x_0$ is in the weak-interior of $X \setminus K$. So $X \setminus K$ is weakly open. Thus $K$ is weakly closed.
  • $\Leftarrow$ Let $K$ be weakly closed. Since the weak topology on $X$ is weaker than the norm topology on $X$ we automatically have that $K$ is norm closed. $\blacksquare$
Corollary 2: Let $X$ be a normed linear space. If $K$ is a norm closed convex subset of $X$ and $(x_n)$ is a sequence in $K$ that weakly converges to $x \in X$ then $x \in K$.
  • Proof: By Mazur's Theorem, since $K$ is a convex subset of $X$ that is norm closed it is also weakly closed. So $K$ is equal to the weak closure of $K$. If $(x_n)$ is a sequence in $K$ that weakly converges to $x \in K$ then since $K = \bar{K}_{\mathrm{weak}}$ we have that $x \in K$. $\blacksquare$
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