Consider the kernel integral operator $\mathcal{K}:L^2[0,1] \mapsto L^2[0,1]$ defined by
$$ \mathcal{K}(f)(t) = \int_0^1 e^{-|t-s|/2}f(s)ds. $$
Is this kernel strictly positive-definite? I.e. for all $f \ne 0$, $\langle \mathcal{K}(f),f \rangle >0$? If so how does one prove such a result? Is anyone aware of a reference containing this result?
1st Solution. Note that
$$ \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{e^{iax}}{1+x^2} \, \mathrm{d}x = e^{-|a|}, \qquad a \in \mathbb{R}. $$
Using this, we get
\begin{align*} \langle \mathcal{K}(f), f \rangle &= \int_{0}^{1} \int_{0}^{1} f(s)f(t) e^{-|s - t|/2} \, \mathrm{d}s\mathrm{d}t \\ &= \frac{1}{\pi} \int_{-\infty}^{\infty} \int_{0}^{1} \int_{0}^{1} f(s)f(t) \frac{e^{i(s-t)x/2}}{1+x^2} \, \mathrm{d}s\mathrm{d}t\mathrm{d}x \\ &= \frac{1}{\pi} \int_{-\infty}^{\infty} \left| \int_{0}^{1} f(s) e^{ixs/2} \, \mathrm{d}s \right|^2 \frac{\mathrm{d}x}{1+x^2}. \end{align*}
This proves that $\mathcal{K}$ is PSD, and moreover, if $\langle \mathcal{K}(f), f \rangle = 0$ then $\int_{0}^{1} f(s) e^{ixs/2} \, \mathrm{d}s = 0$ for every $x \in \mathbb{R}$ (since the integral defines a continuous function in $x$) and hence $f = 0$ in $L^2$ by the Fourier inversion theorem. This shows that $\mathcal{K}$ is strictly PD.
2nd Solution. Let $(X_t)$ be a stationary Ornstein–Uhlenbeck process solving the SDE
$$ \mathrm{d}X_t = -\frac{1}{2} X_t \, \mathrm{d}t + \mathrm{d}W_t, $$
where $(W_t)$ is a standard Wiener process on $\mathbb{R}$. Then $\mathbf{Cov}(X_s, X_t) = e^{-|s-t|/2}$, and so,
\begin{align*} \langle \mathcal{K}(f), f \rangle &= \int_{0}^{1} \int_{0}^{1} f(s)f(t) e^{-|s - t|/2} \, \mathrm{d}s\mathrm{d}t \\ &= \int_{0}^{1} \int_{0}^{1} f(s)f(t) \mathbf{Cov}(X_s, X_t) \, \mathrm{d}s\mathrm{d}t \\ &= \mathbf{Var}\biggl( \int_{0}^{1} f(s) X_s \, \mathrm{d}s \biggr). \\ \end{align*}
This proves that $\mathcal{K}$ is PSD. Moreover, if $\langle\mathcal{K}f, f\rangle$ vanishes, then $C = \int_{0}^{1} f(s) X_s \, \mathrm{d}s$ must be $\mathbf{P}$-a.s. constant with the value
$$ C = \mathbf{E}[C] = \int_{0}^{1} f(s) \underbrace{\mathbf{E}[X_s]}_{=0} \, \mathrm{d}s = 0. $$
Then for any $t \in \mathbb{R}$,
$$ 0 = \mathbf{E}[CX_t] = \int_{0}^{1} f(s) \mathbf{E}[X_t X_s] \, \mathrm{d}s = \int_{0}^{1} f(s) e^{-|t-s|/2} \, \mathrm{d}s. $$
To make use of this observation, note that
$$\mathcal{A} = \{ e^{(s-|t-s|)/2} \}_{t\in[0,1]}$$
is a family of continuous functions on $[0, 1]$ that
So by the Stone–Weierstrass theorem, $\operatorname{span}(\mathcal{A})$ is dense in $C([0, 1])$ w.r.t. the supremum norm, which in turn implies that $\operatorname{span}(\mathcal{A})$ is dense in $L^2$ w.r.t. the $L^2$-norm. Therefore $f(s)e^{-s/2}$ is zero a.e. and $f = 0$ a.e.
The usual tool for working with stationary covariance kernels is the Fourier transform. Let $K : \mathbb{R} \to \mathbb{R}$ be $K(t) = \exp(-|t| / 2)$. Then \begin{align} \langle \mathcal{K}f, g \rangle &= \int_{0}^{1}\int_{0}^{1}K(t - s)f(s)g(t)\,ds\,dt \\ &= \langle K * f, g \rangle \\ &= 2\pi\langle \widehat{K * f}, \widehat{g} \rangle \\ &= 2\pi\langle \widehat{K} \widehat{f}, \widehat{g} \rangle \\ &= 2\pi\int_{-\infty}^{\infty}\hat{f}(\omega)\overline{\hat{g}(\omega)}\hat{K}(\omega)\,d\omega. \end{align} Since $$\hat{K}(\omega) = \frac{1}{2\pi}\int K(x)e^{-i\omega x}\,dx = \frac{2}{\pi(1 + 4\omega^2)}$$ is everywhere positive, the result follows.
