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I have seen this excercise on a book of Functional analysis and I am trying to prove it.

Let $X$ be a vectorial space over a field $\Bbb{K} \, ( \Bbb{R} $ or $ \Bbb{C} )$ and $H \subset X$ a proper subspace. Then, the following are equivalent:

  1. $H$ is maximal (that is, for every subspace $H \subseteq M \subseteq X \rightarrow H=M \lor X=M$ )
  2. It exists a subspace $L \subseteq X$ such that $\dim L = 1, H \cap L = \{ 0 \}$ and $X= H + L $ .
  3. $\dim X/H = 1$
  4. It exists a linear functional $\varphi : X \rightarrow \Bbb{K}$ such that $\ker \varphi = H$

It is important to say that the book does not say anything about the dimension of $X$, in fact there is problem before this about infinite dimensional spaces. This is why I find this problem difficult, the finite case would be very simple.

I saw first that $4 \Rightarrow 3$ is easy to prove using the First Isomorhpy Theorem with $\varphi$ . In order to continue, I have tried to prove $3 \Rightarrow 2$ and $1 \Rightarrow 2$ by considering the linear function $$T: X \times H \rightarrow X, T(x,h) = x-h$$ and taking $L := T( X \times H )$ . This satisfies that $H + L = X$, but I do not know if it works because I am not seeing how to prove $H \cap L = \{ 0 \}$ and $\dim L = 1$ since the possibly infinite dimension of $X$ .

Also, I do not have very idea on how to "reach" $4$ from one of the others. I thought first on consider $\varphi ( x ) = ||| x + H ||| $ where $||| \cdot ||| $ is the norm on $X/H$ induced by a norm on $X$, but I am not sure about it, so any possible help would be appreciated.

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  • $\begingroup$ The $L$ in your attempt cannot work because it equals $X$. $\endgroup$ Apr 2 at 15:59
  • $\begingroup$ This is a purely algebraic theorem. $\Bbb K$ does not need to be $\Bbb R$ or $\Bbb C$, and even if it is and if you have a norm on $X$, $H$ is not supposed to be closed for it. Moreover, the $\varphi$ in your attempt is $\ge0$ hence not linear. $\endgroup$ Apr 2 at 16:02
  • $\begingroup$ Hint for $3\implies1$: for every subspace $H \subseteq M \subseteq X$, $\dim(X/H)=\dim( X/M)+\dim(M/H)$ $\endgroup$ Apr 2 at 16:06
  • $\begingroup$ Hint for $1\implies2$: let $L$ be the line spanned by some $v\in X\setminus H$. Then, $M:=H\oplus L\supsetneq H$ hence $M=X$. $\endgroup$ Apr 2 at 16:09
  • $\begingroup$ Hint for $2\implies4$: if $H\oplus\Bbb Kv=X$, define $\varphi:X\to\Bbb K$ by$$x-\varphi(x)v\in H.$$ $\endgroup$ Apr 2 at 16:13

3 Answers 3

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$1 \implies 2$: Let $v$ be any nonzero vector in $X \setminus H$ and $L = \mathbb{K}v$. Then $L$ is one dimensional and $L \cap H = 0$. Also $H \subsetneq H + L$, which by (1) implies $H + L = X$.

$2 \implies 3$: The conditions $H \cap L = 0$ and $H + L = X$ amount to the stronger $H \oplus L = X$, i.e. every vector $x \in X$ has a unique decomposition as $x = h + \ell$ for $h \in H$, $\ell \in L$. (If you haven't seen direct sums before, try to prove this for yourself!) Thus every coset of $H$ contains a unique element of $L$, namely $x + H \ni \ell$ where the $\ell$ comes from the unique decomposition of $x$. Mapping $x + H \mapsto \ell$ defines a linear map $\bar{f}: X/H \to L$ which is in fact an isomorphism, with inverse sending $\ell$ back to its coset $\ell + H$.

$3 \implies 4$: Since it is one dimensional, choosing any nonzero vector as a basis vector gives an isomorphism $X/H \cong \mathbb{K}$. The quotient map $\phi: X \to X/H \cong \mathbb{k}$ is the desired functional.

$4 \implies 1$: Suppose $H \subseteq M \subseteq X$, and consider the image $f(M) \subset \mathbb{K}$. If $f(M) = \{0\}$, then $M \subseteq \ker(f) = H$, so $M = H$. Alternatively if $f(M) = \mathbb{K}$, then for any $x \in X$ there exists $m \in M$ with $f(x) = f(m)$, subsequently $x - m \in \ker(f) = H \subset M$, so $x = (x-m) + m \in M$ proving $M = X$.

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  • $\begingroup$ Thank you for your answer! $\endgroup$ Apr 3 at 8:31
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$3\implies 2$: the condition $\dim X/H=1$ tells you that there exists nonzero $x_0\in X$ such that for any $x$ there exists a scalar $\lambda$ with $x+H=\lambda x_0+H$. The equality tells you that $x\in\mathbb K x_0+H$.

$2\implies 1$: suppose that $H\subset M\subset X$. By hypothesis any $m\in M$ can be written $m=\lambda x_0+h$ for some $\lambda\in\mathbb K$. If $\lambda=0$ then $m\in H$; and if $\lambda\ne0$, then $x_0=\lambda^{-1}(m-h)\in M$, so $X=M$.

$1\implies 4$: If $H=X$, take $\varphi=0$. Otherwise, let $x_0\in X\setminus H$. By hypothesis, $M=\mathbb K x_0+H=X$. That is, any $x\in X$ can be written as $x=\lambda x_0+h$, with $\lambda\in\mathbb K$ and $h\in H$. This $\lambda$ is unique, for if $\lambda_1 x_0+h_1=\lambda_2 x_0+h_2$, we have $(\lambda_1-\lambda_2)x_0=h_2-h_1\in H$, and so $\lambda_1=\lambda_2$. This allows us to define $$ \varphi(\lambda x_0+h)=\lambda. $$

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  • $\begingroup$ Thank you for your answer! $\endgroup$ Apr 3 at 8:31
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First a few remarks.

  • The $L$ in your attempt cannot work because it equals $X$.
  • This is a purely algebraic theorem. $\Bbb K$ does not need to be $\Bbb R$ or $\Bbb C$, and even if it is and if you have a norm on $X$, $H$ is not supposed to be closed for it. Moreover, the $φ$ in your attempt is $≥0$ hence not linear.

Now, sufficient hints, since you already proved 4⟹3:

  • For 3⟹1: for every subspace $H⊆M⊆X$, $\dim(X/H)=\dim(X/M)+\dim(M/H)$.
  • For 1⟹2: let $L$ be the line spanned by some $v∈X\setminus H$. Then, $M:=H⊕L⊋H$ hence $M=X$.
  • For 2⟹4: if $H⊕\Bbb Kv=X$, define $φ:X→\Bbb K$ by $$x−φ(x)v∈H.$$
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