Dense embedding => dense embedding of dual spaces

Question

Let $E,F$ be Banach spaces, $E\overset{d}{\hookrightarrow}F$ and let E be reflexive. I already proved that $F' \hookrightarrow E'$. Now I want to prove that $F'\overset{d}{\hookrightarrow}E'$.

My first idea was to apply a corollay of Hahn-Banach, namely that $F'\overset{d}{\subset}E'$, if and only if for every $e''\in E''$ and $e'' \mid_{F'}=0$ it follows that $e'' \mid_{E'}=0$.

Let $e'' \in E''$, we know by reflexivity that the canonical mapping $\psi :E \to E''$ (Injection into the double-dual) is bijective, thus there exists a unique $e\in E$ such $\psi(e)=e''$. Now $e'' \mid_{F'}=0 \Leftrightarrow <e'',y'>=0 $ for all $y'\in F'$.

Using the canoncial mapping $0=<e'',y'>=<\psi(e),y'>=<y',e>$ for all $y'\in F'$. Can one conclude that $e=0$ and therefore $e''=0$?

I really appreciate any help you can provide.

No, it is not the restriction operator. If it would be the restriction operator (call it $R$), your first line would contain $ERF$ and this makes no sense... — gerw, Oct 27 '15 at 18:28
I'm not sure what you mean, however I edited my previous post. — fmeyer, Oct 27 '15 at 19:15
Does anybody have a reference with a book stating exactly that result somewhere? — E. Schulz, Mar 25 '21 at 10:37

score 4 · Accepted Answer · answered Oct 28 '15 at 09:38

4

So, we have$\def\<#1>{\left<#1\right>}$, as you write (assuming that $Y'$ is a typo for $F'$), that for all $f' \in F'$: \begin{align*} 0 &= \<e'', f'>_{E',E''}\\ &= \<\psi(e), f'>_{E',E''}\\ &= \<f', e>_{F,F'}\\ \end{align*} Suppose, we had $e \ne 0$, then - by Hahn-Banach - there were $f' \in F'$, such that $f'(e) \ne 0$ (note that $e\in F$), contradicting the above. Hence $e=0$ and therefore $e'' = \psi(0) = 0$.

answered Oct 28 '15 at 09:38

martini

84,101

Does anybody have a reference with a book stating exactly that result somewhere? – E. Schulz Mar 25 '21 at 10:37

Dense embedding => dense embedding of dual spaces

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