Norm of two operators on $l^2(\mathbb{Z}_{2}*\mathbb{Z}_{2})$ different In my research I encounered the following (very concrete) question: Consider the (discrete) group $G:=\mathbb{Z}_{2}*\mathbb{Z}_{2}$. Let $s\text{, }t\in G$ be the generating elements and define for $\theta\in\left(-\frac{\pi}{2}\text{, }\frac{\pi}{2}\right)$ the bounded operator \begin{eqnarray} X_{\theta}:=-8\tan\left(\theta\right)\cdot\text{id}+T_{s}+T_{t}\in{\cal B}\left(l^{2}\left(G\right)\right) \end{eqnarray} on the Hilbert space $l^{2}\left(G\right)$ where $T_{s}\delta_{g}:=\delta_{sg}$ for every $g\in G$ (and $T_{t}$ is defined analogously). Let $P\in{\cal B}\left(l^{2}\left(G\right)\right)$ be the projection onto $\mathbb{C}\delta_{e}$ where $e\in G$ is the neutral element. I claim that \begin{eqnarray} \left\Vert X_{\theta}\right\Vert \neq\left\Vert X_{\theta}-2\tan\left(\theta\right)P\right\Vert \text{,} \end{eqnarray} unless $\theta=0$. At first glance this looks obvious but I could not show it so far.
 A: One can use non-commutative probability theory to compute the spectral distributions of $X_\theta$ and $X_\theta - 2 \tan(\theta) P$ with respect to $\delta_e$.  Since $G = \mathbb{Z}_2 * \mathbb{Z}_2$, the operators $T_s$ and $T_t$ are freely independent with respect to the vector state given by $\delta_e$.  Each of them has the spectral distribution given by the Bernoulli measure $(1/2)(\delta_{-1} + \delta_1)$.  Thus, the spectral distribution of $T_s + T_t$ is the free convolution of two Bernoulli distributions.  Using the $R$-transform (see Chapter 4 of https://arxiv.org/pdf/1908.08125.pdf), we can compute that the free convolution is the measure $\mu$ satisfying
$$
F_\mu(z) = \sqrt{z^2 - 4},
$$
where $F_\mu(z) = 1 / \int (z - t)^{-1}\,d\mu(t)$, also known as the $F$-transform.  Here we use the square root which is defined on the upper half-plane and close to $z$ when $z \to \infty$.  This means that $\mu$ is an arcsine law
$$
d\mu(x) = \frac{1}{\pi \sqrt{4 - x^2}} \chi_{(-2,2)}(x)\,dx
$$
This computation is known; see the last full paragraph at the bottom of page 2 here: https://arxiv.org/pdf/1008.5205.pdf.
Next, let $\nu$ be the spectral measure of $T_s + T_t - 8 \tan(\theta)$.  This just shifts the measure $\mu$ by $-8 \tan(\theta)$, and we have $F_\nu(z) = F_\mu(z + 8 \tan(\theta))$.  Since the support of $\mu$ is $[-2,2]$, the support of $\nu$ is $[-2 - 8 \tan(\theta), 2 - 8 \tan(\theta)]$ and thus the norm of $X_\theta$ is $2 + 8 |\tan \theta|$, at least on the cyclic subspace generated by $\delta_e$.  However, because $\delta_e$ is cyclic for the right shift operators by $s$ and $t$, it is cyclic for the commutant of $X_\theta$ and hence the norm of $X_\theta$ on the entire space agrees with the norm on the cyclic subspace generated by $\delta_e$.
Finally, let's compute the spectral measure of $X_\theta - 2 \tan(\theta) P$, which we will call $\rho$.  This can be done using rank-one perturbation formulas.  Alternatively, $X_\theta$ and $P$ are Boolean independent (see https://mast.queensu.ca/~speicher/papers/boolean.ps).  The spectral measure of $-2 \tan(\theta) P$ with respect to $\delta_e$ is $\delta_{-2 \tan(\theta)}$.  The theory of Boolean independence tells us that
$$
F_\rho(z) - z = (F_\nu(z) - z) + (F_{\delta_{-2\tan(\theta)}}(z) - z) = F_\nu(z) - z + 2 \tan(\theta).
$$
Hence,
$$
F_\rho(z) = \sqrt{(z + 8 \tan \theta)^2 - 4} + 2 \tan(\theta).
$$
Thus, for $z$ in the upper half-plane,
$$
G_\rho(z) = \frac{1}{\sqrt{(z + 8 \tan \theta)^2 - 4} + 2 \tan(\theta)},
$$
where $G_\rho(z) = \int (z - t)^{-1} \,d\rho(t)$.  Using the fact that this is rank-one perturbation that the norm of the new operator $X_\theta - 2 \tan(\theta) P$ is the maximum of the norm of $X_\theta - 2 \tan(\theta) P$ on the cyclic subspace generated by $\delta_e$ and the norm on the orthogonal complement, which will be bounded by $\lVert X_\theta \rVert$. Thus, to show that the two operators have different norms, it suffices to show that the support radius of $\rho$ is strictly larger than that of $\nu$.
Looking at $G_\rho$, the support of the measure $\rho$ agrees with the support of $\nu$ except with the addition of atoms at the point where $(2 \tan \theta)^2 = (z + 8 \tan \theta)^2 - 4$ according to the version of the square root described above.  Assume $\tan \theta > 0$ since the other case is symmetric (and recall $\theta$ was assumed to be in $(-\pi/2,\pi/2)$).  Then we are looking at a point on the negative side where $$(z + 8 \tan \theta)^2 = 4 + 4 \tan^2 \theta = 4 \sec^2 \theta.$$  Hence, $z = -2 \sec \theta - 8 \tan \theta$ is the location of the atom.  Since the support of $\nu$ was already farther on the negative side, we have
$$
\lVert X_\theta \rVert = 2 + 8 |\tan \theta| < 2 \sec \theta + 8 |\tan \theta| = \lVert X_\theta - (2 \tan \theta) P \rVert.
$$
Similar computations can be done for $\mathbb{Z}_2^{*n}$.  Let $T_1$, \dots, $T_n$ be the operators corresponding to the generators.  Then each $T_1 + \dots + T_n$ as a spectral distribution $\mu_n$ is the free convolution of $n$ Bernoulli laws.  This can be done directly with the $R$-transform, or using Proposition 3.1 of this paper: https://arxiv.org/pdf/math/0703295.pdf.  This results in
$$
F_{\mu_n}(z) = \frac{(n-2)z + n \sqrt{z^2 - 4(n-1)}}{2(n-1)}.
$$
Thus, similar computations can be done as in the $n = 2$ case.
Unfortunately, for general groups, it is not as easy to compute because the operators might not be self-adjoint.  However, the tools of free probability could still be used in theory to compute the spectral distribution of $X^*X$.
A: The claim is true.
Any difference in norm must be picked up on the span of $(T_s+T_t)^ne_0.$ So we will apply perturbation theory on that subspace. The value of $\langle(T_s+T_t)^ne_0,e_0\rangle$ should be ${n}\choose{n/2}$ when $n$ is even and zero otherwise. Note that $A=T_s + T_t$ is self-adjoint.  Moreover the spectrum of $A$ contains $2$ and $-2$ as the limits $\|(2+A)^ne_0\|^{1/n}$ and $\|(2-A)^ne_0\|^{1/n}$ are both $4$ by Stirlings type estimates. (In fact, for each $n$ the quantities are equal. This says that the spectral radius of the operators $2+A, 2-A$ are equal to $4.$)
Consider the function
$$F_A(z) = \langle (T_s+T_t-z)^{-1}e_0,e_0 \rangle.$$
The places where $F_A$ analytically continues through $\mathbb{R}$ is exactly the complement of the spectrum.
Expanding $F_A$ at infinity gives:
$$F_A(z) = -\frac{1}{z}\sum {{2n}\choose{n}} \frac{1}{z^{2n}}$$
Now consider $\lim_{z\rightarrow 2^+} F_A(z)$ and $\lim_{z\rightarrow -2^-} F_A(z).$ 
Apparently, using Stirling's formula type estimates, $\lim_{z\rightarrow 2^+} F_A(z)= -\infty.$
Also, as the function is odd,
$\lim_{z\rightarrow -2^-} F_A(z) =\infty.$
By the Aronszajn-Krein formula, the spectrum of 
$A + \alpha P$ is governed by $F_{A+\alpha P}=\frac{F}{1+\alpha F}.$ Note the spectrum will only change if $F(z) = -\frac{1}{\alpha}$ has a real solution in the complement of the spectrum of $A.$ (Moreover, it will only change by one eigenvalue.)
So, now we consider the spectrum of $4\alpha +A$ and compare it to $4\alpha+A + \alpha P.$ If $\alpha >0,$ the extra eigenvalue of $A+\alpha P$ appears when $F_A(z) = -1/\alpha$ which happens to the right of the spectrum, and therefore the norm increases. Similarly, the norm increases in the other case.
Note that it is not true for a general $\alpha + A + \beta P,$ and has a somewhat subtle dependence on your choice of problem.
