In Brezis' "Functional Analysis, Sobolev Spaces and PDEs", the following is stated (Proposition 3.14, page 64):

Let $\varphi:E^\star\rightarrow\mathbb{R}$ be a linear functional that is continuous for the weak$^\star$ topology. Then there exists some $x_0\in E$ such that $\varphi(f)=\langle f,x_0\rangle$ $\forall f\in E^\star$.

The proof goes like this: since $\varphi$ is continous for the weak$^\star$ topology, there are $\varepsilon>0$ and $x_1,\dots,x_n\in E$ such that $$0\in V=\{f\in E^\star:|\langle f,x_i\rangle|<\varepsilon,\forall i=1,\dots,n\},$$ and $|\varphi(f)|<1$ for all $f\in V$. In particular

$(\dagger)\quad$ $\langle f,x_i\rangle =0$ for all $i=1,\dots, n$ implies $\varphi(f)=0,$

and from this, the result follows by a lemma about kernels.

My question is: why does $(\dagger)$ holds?

Thank you!

  • $\begingroup$ What kind of space is $E$? $\endgroup$ Mar 14, 2015 at 7:01
  • $\begingroup$ At the beginning of the section, the author assumes that $E$ is a Banach space. $\endgroup$ Mar 14, 2015 at 7:07

1 Answer 1


If $\langle f,x_i\rangle=0$ for every $i=1,2,...,k$ then for $M>0$, $\langle Mf,x_i\rangle=0$ for every $i=1,2,...,k$. Hence $Mf\in V$ for every $M>0$. But, if $Mf\in V$ then $|\varphi(Mf)|<1$. Finally $|\varphi(f)|<1/M$ for every $M>0$ and hence $\varphi(f)=0$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .