Necessity of parallelogram law for unique existence of projections

Question

Following are two well-known results on taking projections¹ in inner product spaces:

Theorem 1. Any vector of an inner product space has at most one projection on a convex subset.

Theorem 2. Any vector of a Hilbert space can be projected on a nonempty closed convex subset.

Now, the proofs of both the facts hinge crucially on the validity of parallelogram law, and thus fail in general normed (respectively Banach) spaces. Thus, I am looking for examples that show the necessity of the parallelogram law, namely:

1. A normed space in which a convex subset admits two distinct projections of $0$.
2. A Banach space in which $0$ doesn't admit any projection on a nonempty convex closed subset.

Any leads?

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¹ Just to be on the same page, a projection of a vector $x_0$ on a subset $E$ is a vector $y_0\in E$ such that $\|y_0 - x_0\| = \inf_{y\in E}\|y - x_0\|$.

Answer 1

1. Let $X=\mathbb R^2$ with the norm $\|x\|=\max\{|x_1|,|x_2|\}$. Let $$ K=\{x:\ 1\leq x_1\leq 2,\ 0\leq x_2\leq 1\}. $$ Then $K$ is closed, $\def\dist{\operatorname{dist}}$ $\dist(0,K)=1$, and $$ \dist(0,(1,y))=1 $$ for all $y\in [0,1]$.

2. Consider $K\subset C[0,2]$, where $$ K=\{g\in C[0,2]:\ \int_0^1g-\int_1^2g=1\} $$ Then $K$ is closed and convex (it is the preimage of a point by a bounded linear functional). Clearly $0\not\in K$. For $g\in K$, we have \begin{equation}\tag1 1=\int_0^1g-\int_1^2g\leq\int_0^2|g|\leq 2\|g\|_\infty. \end{equation} So $\|g\|_\infty\geq\frac12$ for all $g\in K$.

Now suppose that $\|g\|_\infty=\frac12$. Then both inequalities above are equalities. From the last inequality now turned equality, $$ 0=\int_0^2 (\|g\|_\infty-|g|). $$ As the integrand is non-negative, $|g|=\|g\|_\infty=\frac12$. The first equality in $(1)$ is $$ \int_0^1(\frac12-g)+\int_1^2(\frac12+g)=0. $$ As both integrands are non-negative, this forces $g=\frac12$ on $[0,1]$ and $g=-\frac12$ on $[1,2]$, which is impossible for $g$ continuous.

It remains to show that the distance from $0$ to $K$ is actually $1/2$, that is, that we can find $g\in K$ with $\|g\|$ as close to $\frac12$ as desired. The above reasoning showed that the distance would be achieved by a function that is $1/2$ on $[0,1]$, and $-1/2$ on $[1,2]$; of course this would not be continuous, which is the point. But we can get as close to $1/2$ as follows. Construct $g_n$ as consisting of three segments: namely the lines joining $$ (0,\frac12),\ (1-\frac1n,\frac12+\frac1n),\ (1+\frac1n,-\frac12),\ (2,-\frac12). $$ Explicitly, $$ g_n(x)=\begin{cases} \frac12+\frac1n+\frac 1{n-1}\Big(x-1+\frac1n\Big),&\ 0\leq x\leq 1-\frac1n\\ -\frac12-\frac{n+1}2\Big(x-1-\frac1n\Big),&\ 1-\frac1n\leq x\leq 1+\frac1n\\ -\frac12,&\ 1+\frac1n\leq x\leq 2 \end{cases} $$ Then $g_n\in K$ and $\|g_n\|_\infty=\frac12+\frac1n$.

Answer 2

Here is an example for (1): Take $\mathbb R^2$ with the $\infty$-norm. Take the convex set $\{ (x,y): \ y\ge 1\}$. Then all points $(x,1)$ with $|x|\le 1$ minimize the distance to $0$.