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$Let\ \mathbb{X}$ be a normed space. Its dual space is $\mathbb{X}' := \mathcal{L}(\mathbb{X};\mathbb{K}), $ equipped with the operator norm. Denote $\mathbb{X''}:= \ (\mathbb{X}')'.$ Define the map $\tau : \mathbb{X} \rightarrow \mathbb{X''} $ as follows: for $x \in \mathbb{X}$, $\tau(x)$ is the function from $\mathbb{X'}$ to $\mathbb{K}$ which on input $f \in \mathbb{X}'$ outputs $f(x)$, that is, $\tau(x)(f):=f(x)$.

1. How can I prove this map is well-defined? My understanding is that I should show that for each $x \in \mathbb{X}$, $\tau(x)$ is a linear map $\mathbb{X}' \rightarrow \mathbb{K}$ with a bounded operator norm.

2. Why does the linear map $\tau:\mathbb{X} \rightarrow \mathbb{(X')'}$ have operator norm at most $1$ and how could I show it?

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    $\begingroup$ (1) the map $x\mapsto \tau(x)$, where $\tau(x)(f)=f(x)$ for $f\in X'$ is well defined because $|\tau(x)(f)|=|f(x)|\leq|f||x|=|x||f|$. The inequality is because $f$ is continuous, by definition. $\endgroup$
    – orole
    Jan 31, 2018 at 17:17
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    $\begingroup$ (2) To compute the norm of $\tau$ use the definition. $|\tau(x)(f)|=|f(x)|\leq|x||f|$, therefore $|\tau(x)|\leq |x|=1\cdot |x|$. The $1$ being the best bound because for $|x|=1$ you can find $f$ such that $|\tau(x)(f)|=|f(x)|=1$. It is all confusing, but only because it is so tautological. $\endgroup$
    – orole
    Jan 31, 2018 at 17:21

1 Answer 1

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There is no problem with the definition of $\tau$: for each $x\in X$, you get $\tau(x)\in X''$; the only thing to check is that it is bounded. For that, you have, for any $x\in X$ and $f\in X'$, $$ |\tau(x)f|=|f(x)|\leq\|x\|\,\|f\|. $$ Since $f$ was arbitrary, this shows that $\|\tau(x)\|\leq\|x\|$; and, as $x$ is arbitrary, $\|\tau\|\leq1$.

Now fix $x\in X$, with $\|x\|=1$. By Hahn-Banach, $$ \|x\|=\sup\{|f(x)|:\ f\in X',\ \|f\|=1\}. $$ So for each $\varepsilon>0$ there exists $f\in X'$, with $\|f\|=1$ and $|f(x)|>1-\varepsilon$. Then $$ |\tau(x)f|=|f(x)|>1-\varepsilon, $$ which implies $\|\tau(x)\|>1-\varepsilon$. And this, in turn, implies that $\|\tau\|>1-\varepsilon$. We can do this for all $\varepsilon>0$, so $\|\tau\|\geq1$. Thus $\|\tau\|=1$.

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