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Let $\lambda_j = \tfrac{4}{(2j - 1)^2 \pi^2}$ and $\psi_j(x) = \sqrt{2} \sin(\tfrac{2j-1}{2} \pi x)$ denote the eigenfunctions associated to the first-order Sobolev space.

It is claimed in many texts on reproducing kernel Hilbert spaces that we have $$ \min\{x, y\} = \sum_{j \geq 1} \lambda_j \psi_j(x) \psi_j(y) = \frac{8}{\pi^2} \sum_{n \geq 1, n~\text{odd}} \frac{\sin(\tfrac{n}{2} \pi y) \sin(\tfrac{n}{2} \pi x)}{n^2} $$ for any $x, y \in [0, 1]$.

Equivalently, letting $t = x \pi/2, s = y \pi/2$, $$ \min\{t, s\} = 2 \sum_{n \geq 1, n~\text{odd}} \frac{\sin(n t) \sin(n s)}{n^2} $$ for any $t, s \in [0, \pi/2]$.

Is there a straightforward verification of this fact? I tried applying trigonometric identities but failed to make progress in simplifying the righthand side.

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Drew Brady
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Because of the coefficients $\frac{8}{\pi^2}$ and $\frac{1}{n^2}$ in the series, I got my clue from the Fourier series proof of $$ \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}. $$ After putting things together, here is the most direct solution to your question I have found.

We compute the cosine Fourier series of the function $$ f(x) = 1-x \quad \text{on }[0, 2]. $$ That is, we extend the function evenly to $f(x)=1+x$ on $[-2, 0]$, and then extend the function periodically with period $4$. Then $$ 1-x = a_0 + \sum_{i=1}^\infty a_n \cos \frac{\pi}2 n x, \quad x\in [0, 2]. $$ We compute the Fourier coefficients by standard formulas. \begin{align} a_0 &= \frac{1}{2}\int_0^2 (1-x)\,dx = 0,\\ a_n &= \int_0^2 (1-x)\cos\frac{\pi}2 n x\,dx\\ &= \frac{2}{n\pi}(1-x)\sin\frac{\pi}2 n x \Big|_0^2 + \frac{2}{n\pi}\int_0^2\sin\frac{\pi}2 n x \,dx\\ &=-\frac{4}{n^2\pi^2}\cos\frac{\pi}2 n x\Big|_0^2\\ &=\frac{4}{n^2\pi^2}(1-\cos(n\pi))=\frac{4}{n^2\pi^2}(1-(-1)^n)\\ &=\begin{cases} 0 & n \text{ even},\\ \frac{8}{n^2\pi^2} & n \text{ odd}. \end{cases} \end{align} Therefore, $$ 1-x = \frac{8}{\pi^2}\sum_{n\geq 1,n\text{ odd}} \frac{\cos\frac{\pi}2 n x}{n^2},\quad x\in [0, 2]. $$

Now we adapt this to your case. Without loss of generality, suppose that $x\geq y\in [0, 1]$. Then $\min(x, y)=y$, $x-y, x+y\in [0, 2]$.

By a standard trig identity $$ \sin A\sin B = \frac{1}{2}\big(\cos(A-B)-\cos(A+B)\big), $$ we see \begin{align} &\quad \frac{8}{\pi^2}\sum_{n\geq 1,n\text{ odd}} \frac{\sin(\frac{\pi}2 n x)\sin(\frac{\pi}2 n y)}{n^2}\\ &=\frac 1 2 \frac{8}{\pi^2}\sum_{n\geq 1,n\text{ odd}} \frac{\cos(\frac{\pi}2 n (x-y))-\cos(\frac{\pi}2 n (x+y))}{n^2}\\ &= \frac 1 2 \big((1-(x-y)) - (1-(x+y))\big)\\ &= y = \min(x, y). \end{align} Of course, the formula holds since $x-y, x+y\in [0, 2]$ under the assumption $x\geq y\in [0, 1]$.

If on the other hand $x\leq y\in [0, 1]$, then $x-y\in [-1, 0]$, and the above computation would continue in the last two lines as \begin{align} &=\frac{1}{2}(f(x-y) - f(x+y))\\ &=\frac{1}{2}\big(1+(x-y) - (1-(x+y)\big)\\ &=x = \min(x, y), \end{align} where we use $f(t)=1+t$ when $t\in [-2, 0]$ for $x-y$.

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