2
$\begingroup$

Is there a normed (infinite-dimensional) space $V$ such that the only closed proper subspaces are the finite-dimensional ones?. Without the "closed" hypothesis then it is clearly false (assuming the Axiom of Choice). I think the polynomials in $[0,1]$ may what I'm looking because of the Stone-Weierstrass theorem. Does a positive answer still hold when only talking about Banach spaces?

Thanks in advance

$\endgroup$
3
$\begingroup$

No. On any nonzero normed space $X$, there is (by the Hahn-Banach theorem) a nonzero continuous linear functional $f$. Its nullspace $$ Y := \{x \in X : f(x) = 0\} $$ is a closed linear subspace of $X$. $Y$ is infinite dimensional provided $X$ itself is infinite dimensional.

| cite | improve this answer | |
$\endgroup$
3
$\begingroup$

No. Regarding your example, the polynomials on $[0,1]$ have, among many, the infinite-dimensional closed subspace $$ V_0=\{p:\ p(0)=0\}. $$

For the general case, take $v_0\in V$ with $\|v_0\|=1$. Define a linear functional on $\mathbb C v$ by $\varphi(\lambda v)=\lambda$. Clearly $$|\varphi(\lambda v_0)|=|\lambda|=\|\lambda v_0\|.$$ So Hahn-Banach applies and we can extend $\varphi$ to all of $V$, with $\|\varphi\|=1$. Now define $P:V\to V$ by $$ Pv=\varphi(v)\,v_0. $$ Then $P$ is an idempotent, and $(I-P)V$ is a proper infinite-dimensional subspace of $V$. Moreover, $$ \|Pv\|=|\varphi(v)|\leq\|v\|, $$ so $P$ is bounded. Now suppose that $\{v_j\}\subset (I-P)v$ and $v_j\to v$. Then $$ (I-P)v=\lim_j(I-P)v_j=\lim_jv_j=v, $$ and then $v\in (I-P)V$. So $(I-P)V$ is closed.

| cite | improve this answer | |
$\endgroup$

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.