# crimboi[20] 泛函分析心犯寒

Dieudonne的题, 真实难度不好说, 回顾一下发现其实挺简单的.

##### 题面

Let E be a Hilbert space, F a dense linear subspace of E, distinct from E. (For example E space of l^2 sum of a suitable countable linear independent set and F space of finite sum of this set.) Show that there exists in the prehilbert space F (which simply means viewing F as solely a complex inner product space) a closed hyperplane H s.t. there is no nonzero vector in F that is orthogonal to H.

##### 提示

The construction is stupid: just take some random a\in E-F and H=\lbrace x\in F:\langle x,a\rangle=0\rbrace. It's super easy to show that H is indeed a closed hyperplane of F due to its density. Now \forall y\in F, very naturally consider something like d_H(y), yet d_H is not well defined, so again naturally shift to G=\lbrace x\in E:\langle x,a\rangle=0\rbrace, where d_G is well defined as E is Hilbert. Then comes the ultimate goal showing H is dense in G, which actually follows from some simple computation relying on G being closed and F=H\oplus\mathbb C b for some b\in F-H. I believe then everyone should know what to do finally.

##### 标答

Just prove the density here: as F=H\oplus\mathbb C b and is dense, for any v\in G, we have v_n=\lambda_n b+u_n\rightarrow v where \lambda_n\in\mathbb C and u_n\in H. If \lambda_n\not\rightarrow0, wlog assume |\lambda_n|>\varepsilon>0. As G is closed, \inf\limits_{u\in G}||b-u||>\delta>0 as b\notin H=F\cap G. Then 0=\lim\limits_{n\rightarrow\infty} ||v_n-v||=\lim\limits_{n\rightarrow\infty}|\lambda_n|||b-{u_n-v\over\lambda_n}||\ge\varepsilon\delta>0, contradiction.

THE END