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Problem I'm trying to show that the canonical dual pairing $(\cdot, \cdot):V^{\vee} \times V \rightarrow \mathbb{F}$ between a normed vector space $(V, \mathbb{F})$ and dual $V^{\vee}$ defined by $(f, v) := f(v)$ for all $f \in V^{\vee}$ and $v \in V$ is continuous.

Thoughts The dual paring is a bilinear function; if I can show that $(\cdot, \cdot)$ carries bounded sets to bounded sets, this will suffice since any $n-$linear function that satisfies this criterion is necessarily continuous. It is tempting to write $$ |(f, v)| = |f(v)| \leq |f|_*|v|_V $$ where $|\cdot|_*$ denotes the operator norm on $V^{\vee}$ and $|\cdot|_V$ denotes the norm on $V$. If this equation were true, holding $v$ fixed would provide a bound for $f$ and holding $f$ fixed would provide a bound for $v$. But, the operator norm is only well-defined for bounded linear functions and I'm not certarin that a linear functional is necessarily bounded.

Question So, am I on the right track? If not, how should I approach this problem?

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I guess $V^{\wedge}$ is not the topological dual space. What do we know about $\mathbb F$? – Davide Giraudo Feb 9 '12 at 21:02
@Davide Say, $\mathbb{F}$ is $\mathbb{R}$ or $\mathbb{C}$. The "topological dual"? Does that just mean that $V^{\vee}$ is necessarily the set of all continuous linear forms? If that were part of the hypotheses, I guess that $|\cdot|_*$ is necessarily continuous, and the inequality would hold; correct? – ItsNotObvious Feb 9 '12 at 21:09
Yes by topological dual I mean the set of all linear continuous forms over $V$. Indeed, if $V^{\vee}$ is this set then what you did is correct. In fact, it's the first time I encounter this notation (if it's for the topological dual). It's sometimes denoted by $V'$ or $V^*$. – Davide Giraudo Feb 9 '12 at 21:14
@DavideGiraudo Ok, that was my problem; I missed the fact that the dual space in question was necessarily all continuous forms. The $\vee$ notation is just bad usage on my part. I was thinking of the algebraic dual of $V$ which I usually denote by $V^{\vee}$. If you'll post your comments as an answer I'll accept. Thanks. – ItsNotObvious Feb 9 '12 at 21:20
up vote 1 down vote accepted

What you did is correct: we use the fact that if $E_1,E_2$ and $F$ are three normed spaces, and $b\colon E_1\times E_2\to F$ is a bilinear map then $b$ is continuous if and only if we can find a constant $C$ such that $||b(x_1,x_2)||_F\leq C||x_1||_{E_1}||x_2||_{E_2}$. We apply this result to $E_1=V^*$, the space of all continuous linear forms over $V$ with the usual norm, $E_2=V$ and $F=\mathbb R$ with the absolute value. A constant which works is $C=1$.

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