## Return to Answer

2 edited body

It is pretty obvious after you've seen it, but I like the crinkled curve from Halmos's Hilbert Space Problem book:

Let $f:\mathbb{R}\rightarrow(0,\infty)$ be an $L^2$ function, and define $g\mapsto t\mapsto g_t:\mathbb{R}\rightarrow L^2(\mathbb{R})$ by $$g_t(x)=\chi_{(-\infty,t)}(x) \times f(x).$$

Then $g_t$ has the property that for all $t_1 < t_2 < t_3$ the secants $g_{t_2}-g_{t_1}$ and $g_{t_3}-g_{t_2}$ are mutually orthogonal. (The curve turns a corner at every point.)

1 [made Community Wiki]

It is pretty obvious after you've seen it, but I like the crinkled curve from Halmos's Hilbert Space Problem book:

Let $f:\mathbb{R}\rightarrow(0,\infty)$ be an $L^2$ function, and define $g\mapsto g_t:\mathbb{R}\rightarrow L^2(\mathbb{R})$ by $$g_t(x)=\chi_{(-\infty,t)}(x) \times f(x).$$

Then $g_t$ has the property that for all $t_1 < t_2 < t_3$ the secants $g_{t_2}-g_{t_1}$ and $g_{t_3}-g_{t_2}$ are mutually orthogonal. (The curve turns a corner at every point.)