# “Projection” in normed vector space

Just out of curiosity, is the following true?

If $U$ is a subspace of a normed vector space $X$, and $x\in X\setminus U$, then $$\inf_{u\in U}\|x-u\|>0.$$

If it is true, is it moreover true that there is $u_0\in U$ such that $\|x-u_0\|=\inf_{u\in U}\|x-u\|$?

If the norm is induced by inner product, this is clear because there is a notion of orthogonal projection.

Assuming $\inf_{u\in U}\|x-u\|=0$, then for all $n\in\mathbb{N}$ there exists $u_n$ such that $\|x-u_n\|<\frac1n$. Therefore $\lim u_n=x$. So we get a contradiction if $U$ is closed.

I can't see any obvious contradiction in the general case. My imagination is limited to $\mathbb{R}^n$ $(n=1,2,3)$, so I can't come up with a counterexample (in which $U$ is not closed and hence $X$ is not finite dimensional).

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It is not true in general.

Let $X=\ell_2(\mathbb{R})$, the vector space of square-summable sequences of real numbers, with inner product $$\langle (a_i),(b_i)\rangle = \sum a_ib_i.$$ Let $U$ be the span of the vectors $\mathbf{e}_i$, $i=1,2,\ldots,$, where $\mathbf{e}_i(n) = \delta_{in}$; that is, $\mathbf{e}_i$ is the sequence that is $1$ in the $i$th term, and $0$ elsewhere.

Then $U$ is dense in $X$, but is not closed; since $U$ is dense, for every $x\in X$ we have $\inf_{u\in U}\lVert x-u\rVert = 0$. However, $U\neq X$, as, for example, $(1,\frac{1}{2},\frac{1}{3},\ldots)\in X-U$ (not in $U$ because it is not eventually $0$, and in $X$ because $\sum\frac{1}{n^2}$ is finite).

In fact, the condition $$\inf_{u\in U}\lVert x-u\rVert = 0\implies x\in U$$ is equivalent to saying that $U$ is closed. So your question could be rephrased as asking whether every subspace of a normed vector space is closed (which is why you were able to prove the condition holds if $U$ is closed). This is true in the finite dimensional case, but is not true in the infinite dimensional Hilbert space case (since an infinite dimensional Hilbert space must be of uncountable dimension, but always has a countably dimensional dense subspace given by the span of a Hilbert basis).

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No. For instance, $U$ may be dense in $X$, in which case $\inf_{u \in U} \lVert x-u \rVert = 0$ for all $x \in X$, but the infimum is achieved only for $x \in U$.

One example you may be familiar with: take $X = C([0,1])$, the continuous functions on $[0,1]$, with the supremum norm $\lVert f \rVert = \sup_{t \in [0,1]} |f(t)|$. Let $U$ be the subspace consisting of the polynomials. Then the Weierstrass approximation theorem says that $U$ is dense in $X$, and clearly $U \ne X$.

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