Let $X$ be a Banach space, $M$ be a subspace of $X$ and $x \in X$ be any vector in $X$. Consider $\displaystyle \hat{x}_M=\arg \inf_{m\in M}\|x - m\|$. Under what conditions for $l_p$ norms $p = 1,...,\infty$ is $\hat{x}_M$ unique? Assume that solutions to this infimization problem exist.
Note: Uniqueness may not be possible for all $l_p$ norms. I am aware about counter examples in $l_\infty$ but would like to know counter examples as well as restrictions under which uniqueness would prevail in other $l_p$ norms.
The solutions are unique if the norm of the Banach space is strictly convex: $$ \|\lambda x + (1-\lambda)y\| = \lambda \|x\|+ (1-\lambda)\|y\| $$ only if $x=y$ or $\lambda\in\{0,1\}$. Or equivalently, $$ \|\lambda x + (1-\lambda)y\| < \lambda \|x\|+ (1-\lambda)\|y\| $$ for all $x\ne y$, $\lambda\in(0,1)$.
Now assume that the norm is strictly convex. This implies for $x\ne y$ $$ \|\frac12x + \frac12y\| < \frac12\|x\|+\frac12\|y\|. $$ Take minimizers $x_1,x_2$ of the distance. Assume $x_1\ne x_2$. Then by strict convexity $$ \|\frac12(x_1+x_2)-x\|=\|\frac12(x_1-x) + \frac12(x_2-x)\| < \frac12\|x_1-x\|+\frac12\|x_2-x\| = \inf_{\hat x\in M}\|\hat x-x\|. $$ This is a contradiction and the minimizers are unique. Hence projection is unique in $l^p$, $1<p<\infty$.
The solution need not to be unique in $l^1$: The counterexample works in $\mathbb R^2$ with $\|\cdot\|_1$ norm, it can be extended to $l^1$.
Set $x=(1,1)^T$, $M = \mathrm{span}((1,-1)^T)$. Set $x_1=(0,0)^T$, $x_2 = (-1,1)^T$. Then $$ \|x-x_1\|_1 = 2 = \|x-x_2\|_1. $$ By convexity of the norm, it follows that $x_1$ and $x_2$ are minizers. Hence the projection is not unique in $l^1$.