Is this function represented by a convolution?

Question

I was wondering whether there is an explicit $f$ such that

$\ln(\cosh(x-y)) = \int_{\mathbb{R}} |x-z|f(z-y) dz.$

Why do I believe that this holds?

Because the function $f(a):=\frac{1}{ \cosh^2(a)}$ already seems to come close

see wolframalpha

Answer 1

We can change variables to $z-y$ in the integral and just assume that $y=0$ in what follows.

The answer is in fact no: putting $x=0$ gives $$ \log{\cosh{0}} = \int_{-\infty}^{\infty} \lvert z \rvert f(z) \, dz $$ So we have $$ \int_{-\infty}^{\infty} \lvert z \rvert f(z) \, dz = \int_{0}^{\infty} z (f(z)+f(-z)) \, dz = 0. $$ Differentiating once and evaluating at zero gives $$ 0 = \frac{\sinh{0}}{\cosh{0}} = \int_{-\infty}^{\infty} \frac{z}{\lvert z \rvert} f(z) \, dz = \int_0^{\infty} (f(z)-f(-z)) \, dz. $$ Differentiating twice gives $$ \frac{1}{\cosh^2{x}} = 2f(x) $$ since the derivative of the sign function is $2\delta_0$. Substituting this back in, the second equation is satisfied, but the first is not: $$ \int_{0}^{\infty} \frac{z}{\cosh^2{z}} \, dz = \int_0^{\infty} (1-\tanh{z}) \, dz = \log{2} \neq 0. $$

So you can have $$ \log{2}+\log{\cosh{x}} = \int_{-\infty}^{\infty} \lvert x-z \rvert \frac{1}{2\cosh^2{z}} \, dz, $$ but not without the $\log{2}$: note the convolution operator is linear, but there is no function for which $\int_{-\infty}^{\infty} \lvert x-z \rvert g(z) \, dz = C $: the left-hand side has a nonzero derivative.