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.
Best Answer
It's not surprising that you could not find an example. "Naturally occurring" normed spaces fall into one of two categories:
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).