Suppose $X$ and $Y$ are two anti-Hermitian matrices satisifying $\|X\|, \|Y\| \leq \pi$, where $\|\cdot\|$ is the spectral norm. I'm trying to prove the following bounds on the Frobenius norm of the exponential of them ($e^X, e^Y$ are unitary matrices). Namely, $$ c\|X-Y\|_F \leq \|e^X-e^Y\|_F \leq \|X-Y\|_F, $$ where $c$ is some dimension independent constant, which I believe should be $\frac{2}{\pi}$, but any other constant is okay. And I also suspect (from numerics) that this holds for any unitary invariant norm, not just $\|\cdot\|_F$.
The upper bound follows from triangle inequality over a telescoping sum, which is already figured out: $$ \|e^X-e^Y\|_F \leq \sum_{k=1}^m \|e^{(k-1)X/m}(e^{X/m}-e^{Y/m})e^{(m-k)Y/m}\| = m \|e^{X/m}-e^{Y/m}\|_F, $$ and by taking $m \to \infty$ we have the upper bound.
My question is mainly on the lower bound, which I don't know how to prove. The special case when $X$ and $Y$ commute is easy, because it reduces to proving $$ c\|X\|_F\leq \|e^X-I\|_F. $$ Then we can analyze the eigenvalues directly. Suppose the eigenvalues of $X$ are $i\theta_k$, with $|\theta_k|\leq \pi$. Then for each $\theta_k$, one can prove that $$ |e^{i\theta_k}-1|^2 = 4\sin^2\left(\frac{\theta_k}{2} \right) \geq \frac{4}{\pi^2}\theta_k^2. $$ Thus we have $$ \frac{2}{\pi}\|X\|_F\leq \|e^X-I\|_F. $$ But how to lift this argument to non-commuting $X$ and $Y$ is not clear.
To verify this, I have also generated some random 2-dimensional $X$ and $Y$, and compute the corresponding norms. The result seems to agree with the analysis.
Meanwhile, it is known from Eq. (D3) in this paper that if $\|X\|, \|Y\|\leq r$, then one has the lower bound with $c=2-e^r$. But this is too loose for my purposes. In particular, I would need a lower bound for $\|X\|, \|Y\|\leq \pi$.
Any suggestion is very appreciated. Many thanks in advance!
