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Joseph Van Name
Joseph Van Name
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The weak  $^{*}$ closure of $\{f\in X^{*}:\|f\|=1\}$ is the unit ball $\{f:\|f\|\leq 1\}$$\{f\in X^{*}:\|f\|\leq 1\}$ for infinite dimensional Banach spaces $X$. To prove this, let $f\in X^{*}$ be a functional with $\|f\|\leq 1$. Let $D$ be the set of all finite dimensional subspaces of $X$. Then $D$ is a directed set under inclusion. Let $f_{d}\in X^{*}$ be a linear functional where $f_{d}=f$ on the subspace $D$ and where $\|f_{d}\|=1$. Such an $f_{d}$ exists by a straightforward application of the Hahn-Banach theorem. Take note that for $x\in X$ and $d\in D$ with $x\in d$, we have $f_{e}(x)=f_{d}(x)=f(x)$ whenever $d\subseteq e,e\in D$. Therefore $f_{d}(x)\rightarrow f(x)$. We conclude that $f_{d}\rightarrow f$ in the weak$^{*}$-topology. Therefore, since the set $\{f\in X^{*}:\|f\|\leq 1\}$ is closed in the weak$^{*}$-topology, the set $\{f\in X^{*}:\|f\|\leq 1\}$ is the weak$^{*}$-closure of $\{f\in X^{*}:\|f\|=1\}$.

The weak  $^{*}$ closure of $\{f\in X^{*}:\|f\|=1\}$ is the unit ball $\{f:\|f\|\leq 1\}$ for infinite dimensional Banach spaces $X$. To prove this, let $f\in X^{*}$ be a functional with $\|f\|\leq 1$. Let $D$ be the set of all finite dimensional subspaces of $X$. Then $D$ is a directed set under inclusion. Let $f_{d}\in X^{*}$ be a linear functional where $f_{d}=f$ on the subspace $D$ and where $\|f_{d}\|=1$. Such an $f_{d}$ exists by a straightforward application of the Hahn-Banach theorem. Take note that for $x\in X$ and $d\in D$ with $x\in d$, we have $f_{e}(x)=f_{d}(x)=f(x)$ whenever $d\subseteq e,e\in D$. Therefore $f_{d}(x)\rightarrow f(x)$. We conclude that $f_{d}\rightarrow f$ in the weak$^{*}$-topology. Therefore, since the set $\{f\in X^{*}:\|f\|\leq 1\}$ is closed in the weak$^{*}$-topology, the set $\{f\in X^{*}:\|f\|\leq 1\}$ is the weak$^{*}$-closure of $\{f\in X^{*}:\|f\|=1\}$.

The weak$^{*}$ closure of $\{f\in X^{*}:\|f\|=1\}$ is the unit ball $\{f\in X^{*}:\|f\|\leq 1\}$ for infinite dimensional Banach spaces $X$. To prove this, let $f\in X^{*}$ be a functional with $\|f\|\leq 1$. Let $D$ be the set of all finite dimensional subspaces of $X$. Then $D$ is a directed set under inclusion. Let $f_{d}\in X^{*}$ be a linear functional where $f_{d}=f$ on the subspace $D$ and where $\|f_{d}\|=1$. Take note that for $x\in X$ and $d\in D$ with $x\in d$, we have $f_{e}(x)=f_{d}(x)=f(x)$ whenever $d\subseteq e,e\in D$. Therefore $f_{d}(x)\rightarrow f(x)$. We conclude that $f_{d}\rightarrow f$ in the weak$^{*}$-topology. Therefore, since the set $\{f\in X^{*}:\|f\|\leq 1\}$ is closed in the weak$^{*}$-topology, the set $\{f\in X^{*}:\|f\|\leq 1\}$ is the weak$^{*}$-closure of $\{f\in X^{*}:\|f\|=1\}$.

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Joseph Van Name
Joseph Van Name
  • 1
  • 4
  • 75
  • 118

The weak $^{*}$ closure of $\{f\in X^{*}:\|f\|=1\}$ is the unit ball $\{f:\|f\|\leq 1\}$ for infinite dimensional Banach spaces $X$. To prove this, let $f\in X^{*}$ be a functional with $\|f\|\leq 1$. Let $D$ be the set of all finite dimensional subspaces of $X$. Then $D$ is a directed set under inclusion. Let $f_{d}\in X^{*}$ be a linear functional where $f_{d}=f$ on the subspace $D$ and where $\|f_{d}\|=1$. Such an $f_{d}$ exists by a straightforward application of the Hahn-Banach theorem. Take note that for $x\in X$ and $d\in D$ with $x\in d$, we have $f_{e}(x)=f_{d}(x)=f(x)$ whenever $d\subseteq e,e\in D$. Therefore $f_{d}(x)\rightarrow f(x)$. We conclude that $f_{d}\rightarrow f$ in the weak$^{*}$-topology. Therefore, since the set $\{f\in X^{*}:\|f\|\leq 1\}$ is closed in the weak$^{*}$-topology, the set $\{f\in X^{*}:\|f\|\leq 1\}$ is the weak$^{*}$-closure of $\{f\in X^{*}:\|f\|=1\}$.