A corollary says

Let $F\subset E,$ is a closed linear subspace of $E$, where $E$ is a normed vector space. Then it exists a linear functional $f\in E^*,f\neq 0$ such that $$ \langle x,f \rangle = 0, \forall x\in F .$$

That is, every closed linear subspace $F$ is contained by the kernel of some non-trivial continuous linear functional $f$.

I tried to prove this as follows.

Hahn-Bananch theorem implies for each $x_0\in E \setminus F $, $$\exists f\in E^* \Longrightarrow \langle x,f \rangle \le \alpha \leq \langle x_0, f \rangle, \forall x\in F .$$

But how to preceed ?

Any one helps ?

Thank you in advance

  • 1
    $\begingroup$ It's unclear to me what you're asked to prove. It cannot simply be that every linear subspace of $E$ is contained in some continuous functional's kernel, since the zero functional is always continuous. (Certainly, that's not a corollary of Hahn-Banach.) $\endgroup$ – Jonathan Y. Feb 14 '15 at 22:38
  • $\begingroup$ The triviality pointed out by @JonathanY can, of course, be avoided by requiring $f$ to be non-zero. But then you have to add to the hypothesis that $F$ is not dense in $E$. $\endgroup$ – Andreas Blass Feb 14 '15 at 23:08
  • $\begingroup$ @AndreasBlass do you suggest that the intention was to show that every closed proper subspace of a normed linear space is contained in the kernel of a non-trivial continuous linear functional? $\endgroup$ – Jonathan Y. Feb 15 '15 at 0:16
  • $\begingroup$ @JonathanY. Yes, that's my best guess at a corollary of the Hahn-Banach theorem whose statement is close to the OP's question. $\endgroup$ – Andreas Blass Feb 15 '15 at 0:47
  • $\begingroup$ @AndreasBlass yes, you got what I mean. I have elaborated the question. $\endgroup$ – guanglei Feb 15 '15 at 10:39

Recall the quotient space $E/F$, whose elements are of the form $$[x]:=x+F=\{x+y\in E\,:\,y\in F\}$$ where $x\in E$. This is always a vector space, but given a normed structure on $E$ it is not necessarily obvious how to introduce a related norm on $E/F$. However, if $F$ is closed then we can define a norm as $$\|[x]\|_{E/F}:=\inf_{y\in F}\|x+y\|$$ It is not immediately obvious that this is even a well-defined map, so we shall demonstrate this. Suppose $x,y\in E$ are such that $[x]=[y]$. Then there exists $z\in F$ such that $y=x+z$. Observe that $$\{\|x+w\|\,:\,w\in F\}=\{\|x+z+w\|\,:\,w\in F\}$$ and hence $\|[x]\|_{E/F}=\|[y]\|_{E/F}$, so the map is well-defined. It should be reasonably obvious that $\|\alpha[x]\|_{E/F}=|\alpha|\|[x]\|_{E/F}$, and the triangle inequality is also not too difficult: given $x,y\in E$ and $u,v\in F$ we have $$\|[x+y]\|_{E/F}\le\|x+y+u+v\|\le\|x+u\|+\|y+v\|,$$ so taking infima over all $u,v\in F$ we get $\|[x+y]\|_{E/F}\le\|[x]\|_{E/F}+\|[y]\|_{E/F}$. To show $\|\cdot\|_{E/F}$ is indeed a norm it suffices to show that if $\|[x]\|_{E/F}=0$ then $[x]=[0]$. Suppose $\|[x]\|_{E/F}=0$ and let $(y_n)$ be a sequence in $F$ such that $\|x+y_n\|\to0$. Then $y_n\to-x$ as $n\to\infty$, so since $F$ is closed we have $x\in F$. In particular, this implies $[x]=[0]$. Hence $\|\cdot\|_{E/F}$ is indeed a norm.

Now use the most common corollary to Hahn-Banach: given a normed vector space $X$ and $x\in X$, $x\ne0$ there exists $x^*\in X^*$ such that $\|x^*\|=1$ and $\langle x^*,x\rangle=\|x\|$. Let $[x_0]\in E/F$, $[x_0]\ne[0]$, so by the stated result there is $f\in(E/F)^*$ such that $\|f\|=1$ and $\langle f,[x_0]\rangle=\|[x_0]\|_{E/F}$.

Define $\phi:E\rightarrow\mathbb{K}$ by $\phi(x)=\langle f,[x]\rangle$. Clearly $\phi$ is linear, and $\|\phi(x)\|\le\|f\|\|[x]\|_{E/F}\le\|f\|\|x\|$ so $\phi\in E^*$. If $x\in F$, then $\phi(x)=\langle f,[x]\rangle=\langle f,[0]\rangle=0$. However $\phi\ne0$ since $\phi(x_0)=\|[x_0]\|_{E/F}>0$. This completes our proof.

  • $\begingroup$ not clear for me $\endgroup$ – guanglei Feb 15 '15 at 8:56
  • 1
    $\begingroup$ What exactly isn't clear? I can't really help you if you're not specific. Also, would you mind elaborating in your question to address my first paragraph and the comments? $\endgroup$ – Jason Feb 15 '15 at 9:19
  • $\begingroup$ I have modified the question. $\endgroup$ – guanglei Feb 15 '15 at 13:31
  • $\begingroup$ I have edited the question to include more detail. If you still do not understand I will need you to explain specifically what you are having problems with. $\endgroup$ – Jason Feb 16 '15 at 1:38

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