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Show that all infinite dimensional normed vector space $E$ have a dense hyperplane.

Hint: Consider $\beta$ a convenient Hamel basis of $E$, $S=\mathrm{span}\left\{v_{0},v_{1},\ldots,v_{n},\ldots\right\}$ a countable subset of $\beta$ and let $H=\mathrm{span}\left[\left(\beta\setminus S\right)\cup \left\{\frac{1}{n}v_{n}+v_{0}\:;\:n\geq 1\right\}\right]$.

Remark: I have not been able to build a convenient Hamel base. My attempt was to first build $S$ as a dense, linearly independent and enumerable set, then to extend $S$ to a Hamel basis of $E$, but my attempts have not been successful.

Diego Fonseca
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1 Answers1

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Nobody builds a Hamel basis in an infinite-dimensional space. They just exist, by Zorn's lemma. By "convenient" the hint means that every element of $\beta$ has norm $1$, which is achieved by taking any Hamel basis and normalizing its elements.

Since the kernel of an unbounded functional is dense, it's enough to show there is such a functional $f$. To this end, letting $f(v_n)=n$, and also letting $f(v)=0$ for $v\in\beta\setminus S$, and extending by linearity, would work.

The hyperplane $H$ described in the hint arises in this way if one lets instead $f(v_0)=-1$ and $f(v_n) = n$ for $n\ge 1$ (still $f(v)=0$ for $v\in\beta\setminus S$). I find it a bit more work to do it this way, if one has to show that $H$ is a hyperplane (i.e., the zero set of a linear functional) anyway.

But in case you want to do it this way, observe that

  • $v_0\notin H$, hence $H$ is a proper subspace;
  • if $v_0$ is added to the set that span $H$, the new set spans all of $\beta$, hence all of $H$. Therefore, $H$ is of codimension $1$.
  • I would like to know your opinion about my reasoning to prove that $H$ is a hyperplane: We know that $H$ is a hyperplane of $E$ iff $\mathrm{dim}(E/H)=1$. Then, let $x\in E$, we suppose that $x \notin H$, then there exists $I\subset \mathbb{N}$ finite such that $$x=\sum_{k\in I}\alpha_{k}b_{k}$$ were $b_{k}\in \beta$. We consider $I_{1}$ and $I_{2}$ subsets of $I$ such that $k\in I_{1}$ or $k\in I_{2}$ iff $b_{k}\in \beta\setminus S$ or $b_{k}\in S$ respectively. – Diego Fonseca Oct 30 '17 at 17:17
  • Therefore $$x=\sum_{k\in I_{1}}\alpha_{k}b_{k}+\sum_{k\in I_{2}}\alpha_{k}v_{n_k}=\sum_{k\in I_{1}}\alpha_{k}b_{k}+\sum_{k\in I_{2}}n_{k}\alpha_{k}\left(\frac{v_{n_k}}{n_{k}}+v_{0}\right)-v_{0}\sum_{k\in I_{2}}n_{k}\alpha_{k}.$$ Therefore, calling $\lambda=\sum_{k\in I_{2}}n_{k}\alpha_{k}$ we have $x+H=\lambda(v_{0}+H)$, then $E/H= \mathrm{span}\left{v_{0}+H\right}$, then $\mathrm{dim}(E/H)=1$. We conclude that $H$ is hyperplane. – Diego Fonseca Oct 30 '17 at 17:17
  • Your argument is valid; although it is slightly easier to do it the way I just added at the end of my answer. –  Oct 30 '17 at 17:37