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The James space $J$ is a famous counter-example in functional analysis. It is an example of a Banach space that is isometrically isomorphic to its double dual, but is not reflexive.

Define $$J = \big\{ (a_n) \in c_0\, \big|\, |(a_n)|_J < \infty \big\}$$

where $c_0$ denotes the subspace of $l^{\infty}$ of sequences converging to $0$ and

$$|a_n|_J^2 := \sup\left\{ \sum\nolimits_{i=1}^{k-1} | a_{p_{i+1}} - a_{p_i}|^2 \; \big| \; 1 \leq p_1 < \ldots < p_k \right\} $$

where the supremum is taken over all finite increasing subsequences of $\mathbb{N}$.

How is this different from requiring that $\sum_{n=1}^{\infty}|a_{n+1} - a_n|^2 < \infty$?

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A brief summary of the details of this answer is: consider an appropriately scaled version of the sequence $(0,1,0,\frac14,\frac12,\frac34,1,\frac34,\frac12,\frac14,0,\frac19,\frac29,\dots)$.


These two conditions are different. The James norm being finite is a strictly stronger statement. To see what kind of different behaviour can happen, start by considering the finite sequence $$x = (x_1,x_2,x_3,x_4,x_5) = (0, \frac12, 1, \frac12, 0)$$ Then $\sum_{i=1}^4 |x_i - x_{i+1}|^2 = \sum_{i=1}^4 \frac14 = 1$ but for example, $|x_1 - x_3|^2 + |x_3 - x_5|^2 = 2$ so the James norm is strictly bigger. We can repeat this behaviour to come up with an example illustrating the difference. We'll do this by building suitable blocks with this behaviour.

Let $x^{(k)}_i = \frac{i}{k^2}$ for $0 \leq i \leq k^2$ and let $x^{(k)}_i = 1 - \frac{i}{k^2}$ for $k^2 + 1 \leq i \leq 2k^2$. Then $$\sum_{i=0}^{2k^2 - 1} |x_i^{(k)} - x_{i+1}^{(k)}|^2 = 2k^2 \frac{1}{k^4} = \frac{2}{k^2}$$ (The point of the construction is that the sequence in $k$ on the right hand side is summable). However $|x_0^{(k)} - x_{k^2}^{(k)}|^2 + |x_{k^2}^{(k)} - x_{2k^2}^{(k)}|^2 = 2$.

If we forget for a second that we want $J \subseteq c_0$, this already gives the desired example. Define $$x = (x^{(1)}, x^{(2)}, x^{(3)}, \dots)$$ where I really mean place each block one after the other to build a sequence. The sequence $x$ then goes back and forth between $0$ and $1$ but using smaller and smaller jump sizes. Then $$\sum_{i=1}^\infty |x_i - x_{i+1}|^2 = \sum_{k=1}^\infty \sum_{i=0}^{2k^2 - 1} |x_i^{(k)} - x_{i+1}^{(k)}|^2 = \sum_{k=1}^\infty \frac{2}{k^2} < \infty$$ but $$|x|_J^2 \geq \sum_{k=1}^N |x_0^{(k)} - x_{k^2}^{(k)}|^2 + |x_{k^2}^{(k)} - x_{2k^2}^{(k)}|^2 = \sum_{k=1}^N 2 = 2N$$ for every $N$ and so $|x|_J^2 = \infty$.

The only remaining problem is that $x \not \in c_0$ since it takes the value $1$ infinitely often. This is easily fixed. Let $(a_n)$ be your favourite sequence that is valued in $[0,1]$, converges to $0$ as $n \to \infty$ and is such that $\sum_{n=1}^\infty a_n = \infty$. Define $$y = (a_1^{\frac12} x^{(1)}, a_2^{\frac12} x^{(2)},a_3^{\frac12} x^{(3)}, \dots)$$ where $a_i^{\frac12} x^{(i)}$ means multiply every element of the block $x^{(i)}$ by $a_i^{\frac12}$. Then $y \in c_0$ and $$\sum_{i=1}^\infty |y_i - y_{i+1}|^2 = \sum_{i=1}^\infty a_i|x_i - x_{i+1}|^2 \leq \sum_{i=1}^\infty |x_i - x_{i+1}|^2 < \infty$$ but $$|y|_J^2 \geq \sum_{k=1}^N a_k|x_0^{(k)} - x_{k^2}^{(k)}|^2 + a_k|x_{k^2}^{(k)} - x_{2k^2}^{(k)}|^2 = \sum_{k=1}^N 2a_k \to \infty$$ as $N \to \infty$, so $|y|_J^2 = \infty$.

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With exponent $1$ we have $|a-c| \le |a-b|+|a-c|$. So $$ \sup\left\{ \sum\nolimits_{i=1}^{k-1} | a_{p_{i+1}} - a_{p_i}| \; \big| \; 1 \leq p_1 < \ldots < p_k \right\} = \sum_{n=1}^{\infty}|a_{n+1} - a_n| $$ This means using "finite variation" yields essentially the space $l^1$.
But with exponent $2$, it is false that $|a-c|^2 \le |a-b|^2+|b-c|^2$. Even more, it can happen that $$ \sup\left\{ \sum\nolimits_{i=1}^{k-1} | a_{p_{i+1}} - a_{p_i}|^2 \; \big| \; 1 \leq p_1 < \ldots < p_k \right\} > \sum_{n=1}^{\infty}|a_{n+1} - a_n|^2 $$ in general. (See Rhys example.) So "finite square-variation" yields something quite different than $l^2$.

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