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I want to prove a statement that a subspace $L^m$ of X is of codimension at most m if and only if there exists linear functionals $\lambda_1,\cdots,\lambda_m$ : $X\rightarrow\mathbb{R}$ in the dual space $X^*$ such that $$L^m=\{x\in X:\lambda_i(x)=0\,,\forall i\in[m]\}=\ker A,$$where $A:X\rightarrow\mathbb{R}^m,x\mapsto[\lambda_1(x),\cdots,\lambda_m(x)]^T$

I have an idea to prove "if" part :

$\operatorname{codim}(L^m)=\dim(X)-\dim(L^m)=\dim(X)-\dim(\ker A)$

=$\dim(X)-(\dim(X)-\operatorname{rank}(A))=\operatorname{rank}(A)\leq m$

since A is of size $m\times\dim(X)$

However, now I have no idea regarding the "only if" part

Can anyone help me with the remaining part of this statement? Besides, if my proof of the "if" part has any faults, please also feel free to correct me

Thanks!

Samuel Adrian Antz
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Eric
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1 Answers1

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Assume the codimension of $L^m$ in $X$ is equal $k\le m.$ Let $B_0$ denote the Hamel basis in $L^m.$ By assumptions there are elements $v_1,v_2,\ldots, v_k$ such that the set $B=B_0\cup\{v_1,v_2,\ldots ,v_k\}$ constitutes the Hamel basis in $X.$ Every element $x\in X$ has a unique representation $$x=\sum_{b\in B} x_b b=\sum_{b\in B_0}x_b b+\sum_{j=1}^k x_{v_j}v_j,$$ where only finitely many coefficients $x_b$ are nonzero. Let $\lambda_j(x)=x_{v_j}.$ Then $L_j$ is a linear functional on $X$ and $$L^{m}=\{x\in X\,:\, \lambda_j(x)=0,\ j=1,2,\ldots, k\}$$ If $k=m$ we are done. If not let $\lambda_j=\lambda_k$ for $k<j\le m.$ In this way we get $$L^{m}=\{x\in X\,:\, \lambda_j(x)=0,\ j=1,2,\ldots, m\}$$ If $X$ is a Banach space and we require continuity of the functionals, then the assumption that $L^m$ is closed is necessary. In that case the quotient space $X/L^m$ is a Banach space with norm given by $$[x]=\inf\{\|y-x\|\in L^m\,:\, y\in L^m\}$$ By assumption the quotient space has dimension $k\le m.$ Choose a basis in $X/L^m.$ The elements of the basis are of the form $[v_1],[v_2],\ldots [v_k],$ for some $v_j\in X.$ Every element $[x]$ is uniquely represented as $$[x]=\sum_{j=1}^m \mu_j([x])[v_j]$$ As $X/L^m$ is finite dimensional the functionals $[x]\mapsto \mu_j([x])$ are bounded. Define the functionals $\lambda_j$ on $X$ by $$\lambda_j(x)=\mu_j([x]),\qquad j=1,2,\ldots, m$$ The functionals $\lambda_j$ are bounded as $$|\lambda(x)|=|\mu_j([x])|\le \|\mu_j\|\,\|[x]\|\le \|\mu_j\|\,\|x\|$$ Moreover $$L^m=\{x\in X\,:\, \lambda_j(x)=0,\ j=1,2,\ldots, k\}$$ In case $k<m$ we may repeat the same trick as in the first part.

Ryszard Szwarc
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  • Thank you for your answers! However, I still have some problems : (1) $B_0$ is a Hamel basis for $L^m$ not for X? or how can B and $B_0$ have different dimensions but still be Hamel bases for X simultaneously? (2) The assumptions for "if" part is "at most m" not "equal m", that is, we may assume the codimension of $L^m$ be k ($k\leq m$). Hence, using similar arguments as yours, we can only deduce $L^m = {x\in X:\lambda_j(x)=0, j=1,\cdots, k}$? – Eric Jul 08 '22 at 09:16
  • You are right: $(1)$ was a misprint. For $(2)$ I will re-edit the answer. – Ryszard Szwarc Jul 08 '22 at 10:13
  • Thank you for your supplements! Your answer solved my questions! – Eric Jul 08 '22 at 11:01