Prove that the map $E:V\to V$ is linear.

Question

Let $V$ be an inner product space and $W$ be a finite dimensional subspace of $V.$ Define $E:V\to V$ such that $$E(x)=u$$ where $u$ is the unique best approximation of $x$ on $W$ or in other words $$||x-u||\leq ||x-w||\forall w\in W.$$ I want to show that $E$ is linear. Now my stratergy was to first consider $\gamma\in \mathbb{K}$ and let $E(x)=u.$ We want to show that $E(\gamma x)=\gamma u.$ Since $$||x-u||\leq ||x-w||$$ for any $w\in W$ we have that $$|\gamma||x-u||\leq |\gamma| ||x-w||\implies ||\gamma x-\gamma u||\leq ||\gamma x-\gamma w ||=||\gamma x-w'||.$$

Can I conclude from this that $E(\gamma x)=\gamma u?$

Next, I tried showing that $E(x+y)=E(x)+E(y).$ So first let $E(x)=u$ and $E(y)=u'.$ Then we want to show that $E(x+y)=u+u'.$

Consider $$||x+y-(u+u')||\leq ||x-u||+||y-u'||\leq ||x-w||+||y-w||\forall w\in W.$$

How do I proceed?

Answer 1

I don't know how to prove that $E$ is linear along these lines. Here's a way of doing it (if you are not interested in it, I will delete this answer): take an orthonormal basis $\{e_1,e_2,\ldots,e_n\}$ of $W$ and prove that$$(\forall v\in V):E(v)=\sum_{k=1}^n\langle v,e_k\rangle e_k.$$It is clear then that $E$ is linear.

Answer 2

If you know or can prove that $x-u$ is orthogonal to $W$ (using the initial notations of your post), then you’re done as this proves that $u$ is the orthogonal projection of $x$ on $W$. And the orthogonal projection is a linear map.

Answer 3

First we look for the best approximation of $0$: $$ E(0) = u_0 $$ would give $$ \lVert 0 - u_0 \rVert\le \lVert 0 - w\rVert \quad (w \in W) $$ As $W$ is a subspace of $V$ we have $0 \in W$ and thus $$ \lVert 0-u_0\rVert = \lVert u_0 \rVert \le \lVert 0 - 0 \rVert = 0 $$ and $u_0 = 0$ and $E(0) = 0$ follows.

Then $$ E(\gamma x) = u_{\gamma x} $$ with $$ \lVert \gamma x - u_{\gamma x} \rVert \le \lVert\gamma x - w \rVert \quad (w \in W) $$ and for $\gamma \ne 0$: $$ \lvert \gamma \rvert \lVert x - u_{\gamma x}/\gamma \rVert \le \lvert \gamma \rvert \lVert x - w/\gamma \rVert \quad (w \in W) \iff \\ \lVert x - u_{\gamma x}/\gamma \rVert \le \lVert x - w' \rVert \quad (w' \in W) $$ as $\{w/\gamma \mid w \in W\} = W$, so we see that due to the uniqueness of the best approximation $u_{\gamma x}/\gamma = u$ and $$ E(x) = u = u_{\gamma x} / \gamma = E(\gamma x) / \gamma \iff \\ E(\gamma x) = \gamma E(x) $$ This stays true, if we include $\gamma = 0$, as initially shown.