This R script verifies numerically that Halmos’s crinkled arc is indeed a crinkled arc.

A crinkled arc is a continuous curve in Hilbert space (Hilbert 1982). Specifically, consider \(f\colon \left[0,1\right]\longrightarrow X\), where \(X\) is a Hilbert space with inner product \(\langle\cdot,\cdot\rangle\) . We say that \(f\) is a crinkled arc if it is continuous and possesses the crinkly property: if \(0\leqslant a<b\leqslant c<d\leqslant 1\) then \(\langle f(b)-f(a),f(d)-f(c)\rangle=0\); that is, the chord joining \(f(b)\) to \(f(a)\), and the chord joining \(f(d)\) to \(f(c)\), are orthogonal whenever the intervals \(\left[a,b\right]\) and \(\left[c,d\right]\) are non-overlapping. Halmos makes the observation that every attempt to construct a crinkled arc results in essentially

\[ f(t) = \sqrt{2}\sum_{n=1}^\infty x_n \frac{ \sin\left(n-1/2\right)\pi t }{ \left(n-1/2\right)\pi } \]

where \(x_i\) are an orthonormal set, \(\langle x_i,x_j\rangle=0\) for \(i\neq j\).

We now verify numerically that the Halmos arc does in fact possess the crinkly property, using R and Mathematica. Below, function f() calculates the coefficient of \(x_n\) in the crinkled arc. Because the \(x_i\) are orthogonal, we can calculate \(\left<f(b)-f(a),f(d)-f(a)\right>\) by summing \(f(n,a,b,c,d)\) over \(n=1,2,3,\ldots\).

f <- function(n,a,b,c,d){
  jj <- (n-1/2)*pi
  (sin(jj*b)-sin(jj*a))*(sin(jj*d)-sin(jj*c))/jj^2
}

First a non-overlapping pair of intervals \([0.1,0.2]\) and \([0.3,0.4]\):

plot(cumsum(f(1:100,a=0.1,b=0.2,c=0.3,d=0.4)),type='l')
abline(h=0)

In the above graph, see how the function tends to zero. Now for comparison an overlapping pair \([0.1,0.4]\) and \([0.3,0.5]\):

plot(cumsum(f(1:100,a=0.1,b=0.4,c=0.3,d=0.5)),type='l')
abline(h=0)

In the above graph, the function does not tend to zero.

Mathematica

In mathematica we can try to evaluate the dot product of the two chords symbollically:

Assuming[(a<b) && (b<c)  && (c<d),
      Sum[1/(n-1/2)^2 * (Sin[(n-1/2)*Pi*b]-Sin[(n-1/2)*Pi*a])*(Sin[(n-1/2)*Pi*d]-Sin[(n-1/2)*Pi*c]),
      {n,1,Infinity}
  ]]

but this evaluates to a very complicated and effectively unsimplifiable expression involving a sum of 16 LerchPhi[] functions. However, we can evaluate the expression numerically for a particular set of values of the intervals:

In[1]:= Sum[1/(n-1/2)^2 * (Sin[(n-1/2)*Pi*0.2]-Sin[(n-1/2)*Pi*0.1])*(Sin[(n-1/2)*Pi*0.4]-Sin[(n-1/2)*Pi*0.3]), 
        {n,1,Infinity}] 

                   -12             -17
Out[1]= -2.81775 10    - 2.77556 10    I

(here again \(a=0.1\), \(b=0.2\), \(c=0.3\), and \(d=0.4\)); the value is quite small. For comparison we can choose \(a=0.1\), \(b=0.5\), \(c=0.4\), and \(d=0.8\):

In[2]:= Sum[1/(n-1/2)^2 * (Sin[(n-1/2)*Pi*0.5]-Sin[(n-1/2)*Pi*0.1])*(Sin[(n-1/2)*Pi*0.8]-Sin[(n-1/2)*Pi*0.4]),
        {n,1,Infinity}] 

                            -16
Out[2]= 0.49348 - 1.94289 10    I

which is definitely nonzero.