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 $P \in B(X)$?


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.


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 $$ \|Px\| \le\sum_{i=1}^n\|\phi_i(x)\|\le\sum_{i=1}^n\|\phi_i\|\|x\|. $$


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