Showing there is a projection between a normed space and a subspace

Question

Problem: Let $E$ be a normed space. Suppose $A$ is a finite dimensional subspace of $E$. Show that there exists a continuous projection $T: E \to A.$

Proof. I can write $E=A\oplus B$, where $B$'s existence is given by Zorn, so each $e\in E$ has a decomposition $e=a+b$ with $a\in A$ and $b\in B$. So I would like to define $T(e)=a$. $T$ is necessarily continuous since $A$ is finite dimensional. Or to show cotinuity should I show that $\ker T$ is closed? $T$ is a projection since $T^2(e)=T(a)=a=T(e)$. I am not sure if I am on the right path, this seems too simple.

How does $A$ being finite dimensional guarantee continuity of the projection? — Amitai Yuval, May 20 '15 at 05:37
@AmitaiYuval a linear operator between normed linear spaces continuous iff its bounded? — cap, May 20 '15 at 05:43
The theorem is at least true for Banach spaces by virtue of the Open Mapping Theorem. — Transcendental, May 20 '15 at 05:44
@Transcendental which theorem are you referring to? The statement of the problem or "a linear operator between normed linear spaces continuous iff its bounded"? — cap, May 20 '15 at 05:46
If $ Z $ is a Banach space, and $ X $ and $ Y $ are closed subspaces of $ Z $ such that $ Z = X \oplus Y $, then there exist continuous projections of $ Z $ onto $ X $ and onto $ Y $. In your case, if $ E $ is a Banach space and $ A $ is a finite-dimensional subspace of $ E $, then $ A $ has a closed complement, so the result that I have quoted applies. — Transcendental, May 20 '15 at 05:48
Of course a linear map is continuous if and only if it is bounded, but still, how do you conclude that the projection is bounded? — Amitai Yuval, May 20 '15 at 05:57
By the way, I would define the desired projection by mapping every $e\in E$ to the point in $A$ closest to $e$. — Amitai Yuval, May 20 '15 at 05:59

score 1 · Answer 1 · answered May 20 '15 at 10:47

This is a rather direct solution. Let $\{e_1,\ldots,e_n\}$ be a basis for $A$. Then by linear algebra you might know that $\{f_1,\dots,f_n\}$ is a dual basis for $A^*$ where $f_i(e_j) = \delta_{ij}$. Also, note that $\|f_i\| = 1$ for all $i=1,\ldots,n$. Now by Hahn-Banach Theorem extend $f_i$ to norm one functionals on $E$. Then define an operator $T:E\to A$ by \begin{equation} Tx= \sum_\limits{i=1}^{n}f_i(x)e_i. \end{equation} Now, it is easy to check that $T$ defined above is a projection of $E$ onto $A$. Indeed, for every $y = \sum_\limits{i=1}^{n}\alpha_ie_i \in A$, we have $\alpha_i= f_i(y)$. Therefore, \begin{equation}Ty =\sum_\limits{i=1}^{n}f_i(y)e_i = \sum_\limits{i=1}^{n}\alpha_ie_i = y.\end{equation} Finally, if $x\in X$ and $\|x\|\leq 1$, then \begin{equation}\|Tx\| \leq \sum_\limits{i=1}^{n}|f_i(x)|\|e_i\| \leq \sum_\limits{i=1}^{n}1 = n.\end{equation}

score 1 · Accepted Answer · answered May 20 '15 at 18:51

Let $\{ a_1,\cdots,a_n \}$ be a basis for the subspace $A$, and define a linear map $L : \mathbb{C}^{n} \rightarrow E$ by $$ L(x) = \sum_{k=1}^{n}x_k a_k,\;\;\; x = (x_1,\cdots x_n) \in \mathbb{C}^{n} $$ $L$ is a linear bijection between $\mathbb{C}^{n}$ and $A$. If $\|\cdot\|_{E}$ is the norm on $E$, then $\|x\|_{L} = \|L(x)\|_{E}$ induces a norm on $\mathbb{C}^{n}$. All norms on finite-dimensional spaces are equivalent. Therefore, there are positive constants $m$ and $M$ such that $$ m|x| \le \|x\|_{L} \le M|x|, $$ where $|\cdot|$ is the Euclidean norm on $\mathbb{C}^{n}$. Therefore, the coordinate functionals $F_{k} : A \rightarrow \mathbb{C}$ defined by $F_{k}(x) = (L^{-1}(x))_{k}$ are continuous: $$ |F_{k}(x)| = |(L^{-1}(x))_{k}| \le |L^{-1}(x)| \le \frac{1}{m}\|L^{-1}x\|_{L}=\|x\|_{E}. $$ By the Hahn-Banach Theorem, $F_{k}$ may be extended to a continuous linear functional $\tilde{F_{k}}$ on the entire space $E$. Then $$ Px=\sum_{k=1}^{n}\tilde{F_{k}}(x)a_{k} $$ is a continuous linear operator on $E$ whose range is $A$. And $Pa = a$ for all $a \in A$. Therefore $P^{2}x = Px$, which makes $P$ a projection onto $A$.

Showing there is a projection between a normed space and a subspace

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