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.



Norm of two operators on $l^2(\mathbb{Z}_{2}*\mathbb{Z}_{2})$ different
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.


