Suppose $X$ is a Banach space. If $X$ is separable, then there exists a sequence of functionals $(x^*_n)_{n=1}^{\infty}$ in $X^*$ such that $x_n^*(x) = 0$ for all $n \in \mathbb{N}$ implies that $x = 0.$

Proof: If $X$ is separable, let $(x_n)$ be a sequence of nonzero vectors that is dense in $X$. For each $n$, using the Hahn-Banach theorem, pick $x_n^* \in X^*$ such that $x_n^*(x_n) = \| x_n \|$ and $\| x_n^* \| = 1.$ Suppose $x_n^*(x) = 0$ for all $n$. Then if $\varepsilon>0$, there exists $m \in \mathbb{N}$ such that $\| x-x_m \| < \varepsilon.$ Thus $\| x_m \| = x_m^*(x_m) < \varepsilon,$ and so $\| x \| < 2 \varepsilon.$ Since $\varepsilon >0$ is arbitrary, we have $x = 0.$

I can understand the whole proof, except one part: Why $\| x_m \| = x_m^*(x_m) < \varepsilon$ ? I think we might need the fact $x_n^*(x) = 0$ to conclude the inequality as I do not see it being used anywhere in the proof.


Since $x_m^*(x)=0$, we have

$x_m^*(x_m)=x^*_m(x_m)-x_m^*(x)=x_m^*(x_m-x)\leq |x_m^*(x_m-x)|\leq$

$\leq \Vert x_m^*\Vert \Vert x_m-x \Vert$.

Since $\Vert x^*_m\Vert=1$ and $\Vert x_m-x \Vert<\varepsilon$, by replacing above, one obtains


Since $\Vert x_m\Vert=x_m^*(x_m)$, we finally have

$\Vert x_m\Vert=x_m^*(x_m)<\varepsilon$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.