I've been asked essentially to prove the Hilbert Projection Theorem, but in a general finite dimensional normed vector space (without inner product). The proof I'm familiar with, where you come up with a sequence of vectors converging to the minimum, relies heavily on the polarization identity (and, generally, the existence of an inner product). How exactly would I go about rectifying this without an inner product?

I know that if a vector space $V$ is finite dimensional, then it is isometrically isomorphic to $\mathbb{R}^n$ for some $n$, so we can associate each vector in $V$ with a vector in Euclidean space, and then use the standard inner product, but this seems odd to me.

Alternatively, though this is even more roundabout, that if a normed space satisfies the parallelogram identity, then it's norm induces an inner product: $$ \langle x, y\rangle = \frac14\left(\|x+y\|^2-\|x-y\|^2\right) $$ (which I believe is a theorem of Von Neumann), but I doubt this is what was intended.

Whats the gist of the method for proving this theorem without an inner product? Also, why can't I find this theorem anywhere online? It seems like such a standard result, but searches for "projection theorem on vector spaces" don't seem to turn up anything fruitful.

Edit: To be explicit, I'm trying to prove that for any finite dimensional normed space $V$, and convex subset $X$, there is $x_0 \in X$ that minimizes $\|v - x\|$, for $x\in X$ for a fixed $v\in V$.

  • $\begingroup$ The result will fail in a general Banach space. You need to use the compactness (and convexity) of balls in a finite-dimensional normed space. $\endgroup$ Jun 29, 2018 at 20:04
  • $\begingroup$ $V$ is not isometrically isomorphic to an Euclidean space. They are bi-Lipschitz isomorphic. $\endgroup$
    – user562983
    Jun 29, 2018 at 20:05
  • $\begingroup$ @paulgarrett Thats what I thought! Is the question I'm being asked to prove then simply wrong? $\endgroup$ Jun 29, 2018 at 20:05
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    $\begingroup$ Your convex set $X$ must be assumed to be closed and nonempty, otherwise there need not exist a closest point. $\endgroup$ Jun 29, 2018 at 20:24
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    $\begingroup$ And it's probably implicitly assumed that your $V$ is a real or complex vector space. For $\mathbb{Q}$-vector spaces, just consider $X = \{ x \in \mathbb{Q} : \lvert x\rvert < \pi\}$ and $v$ with $\lvert v\rvert > \pi$. $\endgroup$ Jun 29, 2018 at 20:27

1 Answer 1


Disclaimer: Throughout the discussion, I'll always assume that normed spaces are real or complex (and preferably real).

The existence of some $y$ in $C$ such that $\lVert x-y\rVert=\min_{z\in C} \lVert x-z\rVert$ is a consequence of the fact that closed balls are compact. There must be some closed ball $E(x,R)$ such that $E(x,R)\cap C\ne \emptyset$. Since $C$ is closed and $E(x,R)$ is compact, $E(x,R)\cap C$ is compact; therefore, there is some $y\in E(x,R)\cap C$ that minimizes the continuous function $\lVert x-\bullet\rVert$ on $E(x,R)\cap C$. Now, it is easy to observe that $\min\left\{\lVert x-z\rVert\,:\, z\in C\right\}=\min\left\{\lVert x-z\rVert\,:\, z\in C\cap E(x,R)\right\}$. For, if $z\notin E(x,R)$, then $\lVert x-z\rVert$ is automatically larger than $R$ and thus of the distance from $x$ of any element of $E(x,R)\cap C$.

So $y$ is indeed a minimizer such as the ones you want. Without additional hypothesis uniqueness won't be guaranteed, basically for the same reason it isn't guaranteed in general. If $V=(\Bbb R^2,\lVert\bullet\rVert_\infty)$ and $C=\{1\}\times [-1,1]$ and $x=(0,0)$, then $d(x,y)=1$ for all $x\in C$.

Following this line of thought, you may want to prove that a normed space is strictly convex if and only if it has the property that for all closed convex sets $C$ and $x\notin C$ there is at most one $y$ such that $\lVert x-y\rVert=\min\{\lVert x-z\rVert\,:\, z\in C\}$.


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