Examples of strictly convex normed spaces that are not uniformly convex I am studying approximation theory, just to learn basically, and in different books I saw a theorem about uniqueness of best approximation. One book uses strict, the other uses uniform convexity with the same result and proof steps. And I know that uniform convexity implies strict convexity.
So obviously if theorem works for strictly convex then it works for uniformly convex. And I tried to find some examples of such spaces, but I could not find one that is strictly convex (but not uniformly) normed linear space.
Any examples you can give is appreciated.
 A: It's not surprising that you could not find an example. "Naturally occurring" normed spaces fall into one of two categories: 


*

*Very nice: uniformly convex and uniformly smooth 

*Not nice at all: not strictly convex, not smooth (not reflexive, etc). 


However, one can artificially construct a norm that is strictly convex without being uniformly convex. Here is one: on $\ell^1$, the space of absolutely summable sequences, 
let $$\|x\| = \sum_{n=1}^\infty |x_n| + \sqrt{\sum_{n=1}^\infty |x_n|^2}\tag1$$
The second sum makes the space strictly convex, since it's the Hilbert space norm. However, the first sum dominates and determines the topology of the space. Indeed, we have 
 $$\sum_{n=1}^\infty |x_n|\le \|x\| \le 2\sum_{n=1}^\infty |x_n|  $$
So, the new norm is equivalent to the original norm of $\ell^1$. Consequently, it is not uniformly convex: uniform convexity implies reflexivity, and $\ell^1$ is not reflexive. (The property of being reflexive is preserved when a norm is replaced by an equivalent one.)
Here is a direct way of seeing that (1) is not a uniformly convex norm. Let $x_n=1$ for $n\le N$ and $=0$ otherwise. Let $y_n=1$ for $N<n\le 2N$ and $=0$ otherwise. Then $\|x\|=\|y\|=N+\sqrt{N}$ and $\|x-y\| \ge 2N$ while 
$$\left\|\frac{x+y}{2}\right\| \ge N   
$$
Dividing both vectors by $N+\sqrt{N}$, we see uniform convexity failing.

By the way: while strict convexity is enough for uniqueness of best approximation, it may not be enough for existence. Uniform convexity gives both uniqueness and existence (if the space is complete).  
