Set the elements in corners to keep the arrow length same in a tikzcd commutative diagram

Let $W_2(mu,nu)$ denote the $2$-Wasserstein distance between two given probability measures $mu$ and $nu$ on $mathbb R^n$. For a probability measure $mu$ and $f:mathbb R^nto mathbb R^n$, let $f_{#}mu=mucirc f^{-1}$ denote the push-forward of $mu$ under $f$, i.e. $(f_{#}mu)(B)=mu(f^{-1}(B))$ for every Borel set $B$ in $mathbb R^n$. Why does the following inequality hold true?
$$W^2_2(f_{#}mu,g_{#}mu)leq int_{mathbb R^n}|f(x)-g(x)|^2,dmu(x) $$
for all $mu$-measurable functions $f,g:mathbb R^ntomathbb R^n$.

Some comment: the product measure $f_{#}muotimes g_{#}mu$ is a so-called transport plan and by definition of the Wasserstein distance
$$W^2_2(f_{#}mu,g_{#}mu)leq int_{mathbb R^ntimes mathbb R^n}|x-y|^2,d(f_{#}muotimes g_{#}mu)(x,y)=int_{mathbb R^ntimes mathbb R^n}|f(x)-g(y)|^2,dmu(x),dmu(y).$$

I am currently studying construction of the Grothendieck group of a commutative monoid $M$. I was looking for an example of a monoid that is torsion, namely, I have the following query.

Does there exist a monoid $M$ (written multiicatively with identity element denoted by $1$) that is not a group such that $x in M$ implies $x^n = 1$ for some $n in mathbb{N}$?

Generating functions seem to be a powerful tool in discrete mathematics for solving differential equations and recurrence relations. I’ve been trying to figure out if these methods can be expanded to differential equations that involve matrices, such as Schrodinger’s equation. Is there anything that prevents a solution to these types of differential equations being written using generating functions? For example, given a solution to Schrodinger’s equation as

$$ U = Te^{-i hbar int_{0}^{t’} H(t) ,dt} $$

Where H(t) is some time-dependent Hamiltonian matrix and T is the time-ordering operator. This form seems very reminiscent to exponential generating functions. For example, the generating function for the Bessel functions is given by $$ e^{frac{x}{2}(t-frac{1}{t})}=sum_{n=-infty }^{infty} J_n(x)t^n $$

So, if the time evolution of the Hamiltonian gave something similar to the generating function above, where x is now the matrix Hamiltonian, can the exponential be re-written in a form using the Bessel functions of matrix argument? I’d assume that the matrix argument of a Bessel function would be similar to an analytic function of matrix argument, but everything I find seems to write Bessel/Hypergeometric functions in terms of zonal polynomials, and the wikipedia page for hypergeometric functions of matrix argument even mention that these types of functions aren’t similar to writing other functions in terms of matrix arguments. That doesn’t make sense to me though since these special functions have these generating function relations. Any help would be appreciated if someone could point me towards the literature too.

I am trying to find the $L$-smoothness constant of the following function (logistic regression cost function) in order to run gradient descent with an appropriate stepsize.

The function is given as $f(x)=-frac{1}{m} sum_{i=1}^mleft(y_i log left(sleft(a_i^{top} xright)right)+left(1-y_iright) log left(1-sleft(a_i^{top} xright)right)right)+frac{gamma}{2}|x|^2$ where $a_i in mathbb{R}^n, y_i in{0,1}$,$s(z)=frac{1}{1+exp (-z)}$ is the sigmoid function.

The gradient is given as
$nabla f(x)=frac{1}{m} sum_{i=1}^m a_ileft(sleft(a_i^{top} xright)-y_iright)+gamma x $.

My ideas was that the smoothness constant $L$ has to be bigger than all the eigenvalues of the hermitian of the given function, this follows from the fact that if $f$ is $L$-smooth, $g(x)=frac{L}{2} x^T x-f(x)$ is a convex function and therefore the hessian has to be positive semi-definite.
The second-order partial derivatives of $f$ are given as

$ frac{partial^2 }{partial x_k partial x_j}f(x)=frac{1}{m} sum_{i=1}^ms(a_i^{top} x)left(1-s(a_i^{top} x)right)[a_i]_k[a_i]_j+gammadelta_{ij} $

from the following github post (https://github.com/ymalitsky/adaptive_GD/blob/master/logistic_regression.ipynb) i know that $ L=frac{1}{4} lambda_{max }left(A^{top} Aright)+gamma$ , where $lambda_{max }$ denotes the largest eigenvalue, which seems good since i figured out that $s(a_i^{top} x)left(1-s(a_i^{top} x)right)leq frac{1}{4}$ for all $x$.

But i am not able to fit everything together. I would appreciate any help.