3
$\begingroup$

Let $X$ be a normed linear space and $X_0$ be a finite dimensional subspace of $X$. Prove that there exists a projection operator $P \in B(X)$ such that $R(P ) = X_0$.

My Try: $X$ can always be written as $X_1 + X_0$ where $X_1$ is a closed subspace and $X_0 \cap X_1 = \{0\}$. So $x \in X, x = y+z, y \in X_0, z\in X_1.$ Define the map $P(x)=y$. I think that this map works.

Can someone check the argument is correct or not? And how to show $P \in B(X)$?

$\endgroup$
5
$\begingroup$

Let $X_0 \le X$ be a finite-dimensional subspace of $X$ and let $\{e_1, \ldots, e_n\}$ be a basis for $X_0$. Then every $x \in X_0$ has a unique representation as a linear combination of $e_1, \ldots, e_n$:

$$x = \alpha_1(x)e_1 + \ldots \alpha_n(x)e_n$$

Notice that $\alpha_i : X_0 \to \mathbb{F}$ are linear functionals, and they are bounded because since functionals on a finite-dimensional space are always bounded.

Thus, using the Hahn-Banach theorem, we can extend them to bounded functionals $\tilde\alpha_i : X \to \mathbb{F}$.

Now define $P : X \to X_0$ as

$$Px = \tilde\alpha_1(x)e_1 + \ldots \tilde\alpha_n(x)e_n, \quad x\in X$$

Notice that $P$ acts as identity on $X_0$:

$$Px = \tilde\alpha_1(\underbrace{x}_{\in X_0})e_1 + \ldots \tilde\alpha_n(\underbrace{x}_{\in X_0})e_n = \alpha_1(x)e_1 + \ldots \alpha_n(x)e_n = x, \quad x \in X_0$$

Therefore, $\DeclareMathOperator{\Ima}{Im}$$P^2 = P$ since $\Ima P = X_0$.

Finally, $P$ is bounded:

\begin{align} \|Px\| &= \|\tilde\alpha_1(x)e_1 + \ldots \tilde\alpha_n(x)e_n\| \\ &\le \|\tilde\alpha_1(x)\|\|e_1\| + \ldots \|\tilde\alpha_n(x)\|\|e_n\| \\ &\le \|\tilde\alpha_1\|\|x\|\|e_1\| + \ldots \|\tilde\alpha_n\|\|x\|\|e_n\| \\ &= \big(\|\tilde\alpha_1\|\|e_1\| + \ldots \|\tilde\alpha_n\|\|e_n\|\big)\|x\| \end{align}

Thus, $P$ is the desired projection.

From $\DeclareMathOperator{\Ker}{Ker}$here we also get the decomposition $X = X_0 \dot+ \Ker P = X_0 \dot+ \bigcap_{i=1}^n \Ker\tilde\alpha_i$. Thus follows that a finite-dimensional subspace always has a direct complement, a fact which cannot be used before conducting a proof similar to this one.

$\endgroup$
2
$\begingroup$

It's not true that $X$ can always be decomposed in this way for any given closed subspace. It's critical that $X_0$ is finite dimensional.

Since $X_0$ is finite dimensional, it is isomorphic to a Hilbert space of finite dimension, say $n$. If $e_1,\ldots,e_n$ is an orthonormal basis for $X_0$ with the inner product $\langle \cdot,\cdot\rangle$, then the mapping $x\mapsto \langle x,e_i\rangle$ is a bounded linear functional on $X_0$ for each $i=1,\ldots,n$. Using the Hahn-Banach theorem we can extend each of these to a bounded linear functional $\phi_i$ on $X$. Now define $P:X\to Y$ by $P(x)=\sum_{i=1}^n\phi_i(x)e_i$. Since every $x\in X_0$ satisfies $x=\sum_{i=1}^n\langle x,e_i\rangle e_i=P(x)$, we infer that $P$ is a projection onto $X_0$. Moreover, $P$ is bounded because $$ \|Px\| \le\sum_{i=1}^n\|\phi_i(x)\|\le\sum_{i=1}^n\|\phi_i\|\|x\|. $$

$\endgroup$

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.