Sutter et al. [1] in their paper "Multivariate Trace Inequalities" give an intuitive proof of the following Golden-Thompson inequality:


For any hermitian matrices $A,B$:


$$
\text{tr}(\exp{(A+B)}) \le \text{tr} \exp{(A)}\exp{(B)}.
$$


Lemmas involve the spectral pinching method which uses the eigendecomposition $A=\sum\_{\lambda}\lambda P\_\lambda$ where the $\lambda$ are eigenvalues and $P\_\lambda$ corresponding projectors which are mutually orthogonal.


The spectral pinching map with respect to $A$ is defined as
$$
\mathcal{P}\_A: X \mapsto \sum\_{\lambda} P\_\lambda XP\_\lambda
$$
and then come these properties for any $X \geq 0$:


1) $\mathcal{P}\_A[X]$ commutes with $A$ 


2) $\text{tr} \mathcal{P}\_A[X]A = \text{tr} X A$ 


3) \begin{align}
\mathcal{P}\_A[X] &= \sum\_{\lambda \in \text{spec}(A)} P\_\lambda XP\_\lambda\\
&= \frac{1}{|\text{spec}(A)|} \sum\_{y=1}^{|\text{spec}(A)|}U\_yXU\_y^\*\\
&\geq \frac{1}{|\text{spec}(A)|} X
\end{align}


where $\text{spec}(A) = \{\lambda\_1, \lambda\_2, \dots, \lambda\_{|\text{spec}(A)|}\}$ 
and $U\_y = \sum\_{z=1}^{|\text{spec}(A)|} \exp{\frac{i2\pi yz}{|\text{spec}(A)|}}P\_{\lambda\_z}$ satisfies $UU^T=I$


1) and 2) are straightforward to follow. But I cannot understand how the second equality and the first inequality hold true for 3).


[1] <https://arxiv.org/abs/1604.03023>



Intuitive proof of Golden-Thompson inequality
$\newcommand{\al}{\alpha}
\newcommand{\la}{\lambda}$
Let $\la\_1,\dots,\la\_n$ be the distinct eigenvalues of $A$, so that $|\text{spec}(A)|=n$. Then 
\begin{multline}
\sum\_{y=1}^n U\_yXU\_y^\*=\sum\_{y=1}^n\sum\_{u,v=1}^n e^{i2\pi yu/n}P\_{\la\_u}XP\_{\la\_v}e^{-i2\pi yv/n}\\
=
\sum\_{u,v=1}^n P\_{\la\_u}XP\_{\la\_v}\sum\_{y=1}^n e^{i2\pi y(u-v)/n} 
=\sum\_{u,v=1}^n P\_{\la\_u}XP\_{\la\_v}n1\_{\{u=v\}} 
 =n\sum\_{u=1}^nP\_{\la\_u}XP\_{\la\_u},
\end{multline}
so that the 2nd equality in 3) holds. 


Now, as pointed out in the cited paper, $U\_yXU\_y^\*\ge0$ for all $y$, whereas $U\_n=I$. So, the inequality in 3) follows as well. 


