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.

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!