Is there an easy example of a strictly-convex space that is not uniformly convex? I've found on the web that $\ell^1$ equipped with the norm $\|\cdot \|=\|\cdot\|_1+\|\cdot\|_2$ is an example. However, I can't prove this to see that it is true. Can someone show me how, or provide a different, and preferably, easier example. Another example that I found was related to the space $c_0$ with a different complicated norm, which again is difficult to prove.
 A: The space $\ell^1(\mathbb{N})$ endowed with the norm $\lVert x\rVert = \lVert x\rVert_1 + \lVert x\rVert_2$ is a simple example, and off the top of my head, I can't think of a simpler one.
To see that the space is strictly convex, it is convenient to use the last characterisation of strict convexity listed on the wikipedia page:

A Banach space $(V,\lVert\,\cdot\,\rVert)$ is strictly convex if and only if $x\neq 0,\, y \neq 0$ and $\lVert x+y\rVert = \lVert x\rVert + \lVert y\rVert$ together imply that $x = cy$ for some constant $c > 0$.

Rather, it is convenient to use the equivalent characterisation that a Banach space is strictly convex if and only if $\lVert x+y\rVert < \lVert x\rVert + \lVert y\rVert$ for all linearly independent $x,y$.
It is clear that the former characterisation implies $\lVert x+y\rVert < \lVert x\rVert + \lVert y\rVert$ for all linearly independent $x,y$, and for the converse, we note that $\lvert 1+c\rvert = 1 + \lvert c\rvert$, where $c \in \mathbb{C}$, holds if and only if $c$ is real and non-negative.
This characterisation makes it easy to see that if $\lVert\,\cdot\,\rVert_s$ is a strictly convex norm and $\lVert\,\cdot\,\rVert_a$ is an arbitrary norm, then $\lVert\,\cdot\,\rVert_a + \lVert\,\cdot\,\rVert_s$ is strictly convex, for in the triangle inequality
$$\lVert x+y\rVert_a + \lVert x+y\rVert_s \leqslant \lVert x\rVert_a + \lVert y\rVert_a + \lVert x\rVert_s + \lVert y\rVert_s,$$
the inequality for $\lVert\,\cdot\rVert_s$ is strict when $x$ and $y$ are linearly independent.
Since $\lVert\,\cdot\,\rVert_2$ is uniformly convex, and hence strictly convex, it follows that $(\ell^1(\mathbb{N}), \lVert\,\cdot\,\rVert)$ is strictly convex.
It remains to see that $\lVert\,\cdot\,\rVert$ is not uniformly convex. For $n \geqslant 1$, let
\begin{align}
a_n &= \sum_{k = 1}^n e_{2k},\\
b_n &= \sum_{k = 1}^n e_{2k+1}.
\end{align}
Then $\lVert a_n\rVert_1 = \lVert b_n\rVert_1 = n$ and $\lVert a_n\rVert_2 = \lVert b_n\rVert_2 = \sqrt{n}$, so $x_n = \frac{1}{n + \sqrt{n}}a_n$ and $y_n = \frac{1}{n+\sqrt{n}}b_n$ are sequences of norm $1$, and
$$\lVert x_n + y_n\rVert = \frac{1}{n + \sqrt{n}} \lVert a_b + b_n\rVert = \frac{1}{n + \sqrt{n}} (2n + \sqrt{2n}) \to 2,$$
but also $\lVert x_n - y_n\rVert = \lVert x_n + y_n\rVert > 1$, so the norm isn't uniformly convex.
