Showing that a best approximation exists in a finite-dimensional subspace?

Question

I have some questions on a proof that for every element of a normed space $X$, there exists a best approximation of this element in a finite-dimensional subspace $U$. Here is the proof:

Let $\varphi \in X$ and choose a minimizing sequence $(u_n)$ for $\varphi$, i.e. $u_n \in U$ satisfies $$ \||\varphi-u_n|| \to d:= \inf_{u\in U} ||\varphi-u||, \quad n\to \infty. $$ As $||u_n|| \le ||\varphi-u_n|| + ||\varphi||$ the sequence $(u_n)$ is bounded. Since $U$ has finite dimension, it is closed, and hence the sequence $(u_n)$ contains a convergent subsequence $(u_{n(k)})$ with limit $v\in U$. Then we have that $$ ||\varphi-v|| = \lim_{k\to\infty} ||\varphi-u_{n(k)}|| = d. $$

Questions:

1. Why do we have to go to the trouble of starting with a minimizing sequence? Say $X=\mathbb{R}^2$, $U = \{(x,0):x\in\mathbb{R}\}$, and we take the point $\varphi = (5,2)\in \mathbb{R}$. Then the best approximation to $\varphi$ in $U$ is clearly $u=(5,0)$. So why do we need minimizing sequences?
2. Why does $||u_n|| \le ||\varphi-u_n|| + ||\varphi||$ mean that $(u_n)$ is bounded? The first term on the right hand side could be arbitrarily large, so how does this give us a bound $M$ such that $||x_n|| \le M$ for all $n\in \mathbb{N}$?
3. Why are we not done after the first equation? That is, we defined a minimizing sequence in $U$ that converges to the best approximation..so why can't we take the limit of the sequence there and then and be done with it?

Answer 1

1) If you want to use closed + bounded $\implies$ convergent subsequence, then you have to start with a sequence. But this is a weasely way to address your question. Here's a better way.

Suppose there were no best approximation in $U$. Then there is surely a sequence of successively improving approximants in $U$. (Otherwise: "Hey, look! We found the best approximant.") The rest of the proof goes on to show that the resulting sequence has an eventually constant subsequence. (It's worth pointing out there can be multiple best approximants and our starting sequence can eventually settle down to just output these "with some random pattern of repetition". But then at least one of them is repeatedly infinitely often, so there is a convergent subsequence.)

If you start by writing down (one of) the best approximants, then you are of course done. How do you do this generically and simultaneously for all finite dimensional subspaces of all normed spaces? That's trickier. Are we even sure that sequences of improving approximants ever settle down in unfamiliar spaces and/or with unfamiliar norms?

2) "$||\phi - u_n|| \longrightarrow d$" means for any $\varepsilon > 0$, there exists a positive integer $N$, such that for all $n > N$, $||\phi - u_n|| \in (d- \varepsilon, d + \varepsilon)$. Consequently, the first term on the right hand side is bounded by, say, $d + 1$ for sufficiently large $n$. So the right hand side is bounded by $||\varphi|| + d + 1$. (In fact any number $||\varphi|| + d + \varepsilon$ for $\varepsilon > 0$ is established as an upper bound by this argument.)

3) After the first equation, I have an infimum on the right-hand side. There's no reason to believe that the infimum is actually attained by any $u \in U$. (In fact, this is what we are trying to prove: that there is a $u$ for which the infimum is attained.) Also, I have a limit on the left-hand side. Again, there's no reason to assume that there is a $u_n$ that makes "$\rightarrow$" actually "$=$". So you have a sequence of $u_n$ that can get you as close as you like to $d$ and you know there are elements of $U$ that can get you as close as you like to $d$ from above. How do you know that you can ever actually attain $d$?

(It's counterfactual for this problem, but could be a useful example of how the idea your third question suggests doesn't quite work. Suppose the limit on the left-hand side were only approached from below. Then the upper bound for the left-hand side would match the lower bound for the right-hand side. But that wouldn't tell you that any value appearing in the sequence on the left agreed with any value in the set on the right. Unless both actually attain their bounds. So there's more to do.)

Answer 2

1. You want a proof with no assumption on what $X$ and $U$ could be.

2. A convergent sequence is bounded.

3. From the first equation you don't know how $u_n$ behaves, nor do you have any candidate for a best approximation.