r^{k+1}\right\rVert_{2}+s_{k}\left\lVert\epsilon_{1}^{k}\right\rVert_{2}+C_{\rho},$

where $C_{\rho}=\sqrt{2Lc_{2}\rho}+c_{1}L\rho$. If $E^{k+1}:=\left\lVert
r^{k+1}\right\rVert_{2}+s_{k}\left\lVert\epsilon_{1}^{k}\right\rVert_{2}$ is a
positive and absolutely summable sequence, then $\\{x^{k}\\}_{k\geq 0}$ is a
quasi-Féjer sequence.

###### Proof.

Writing $\text{prox}^{\epsilon_{2}^{k}}_{s_{k}h}(x)$ as
$\text{prox}_{s_{k}h}(x)+r^{k}$ and $\nabla^{\epsilon_{1}^{k}}g(x)$ as $\nabla
g(x)+\epsilon_{1}^{k}$ for any optimal solution $x^{\star}$ of (10), we obtain

(186)
$\begin{split}\left\lVert{x}^{k+1}-x^{\star}\right\rVert_{2}&=\bigg{\|}\text{prox}_{s_{k}h}(x^{k}-s_{k}\nabla
g(x^{k})-s_{k}\epsilon_{1}^{k})\\\
&\quad-\text{prox}_{s_{k_{0}}h}(x^{\star}-s_{k_{0}}\nabla
g(x^{\star})-s_{k_{0}}\epsilon_{1}^{k_{0}})+r^{k+1}-r^{k_{0}}\bigg{\|}_{2}.\end{split}$

By assumption we have $\epsilon_{2}^{k}\leq
c_{2}\left\lVert{x}^{k+1}-{x}^{k}\right\rVert_{2}$, or equivalently,
$\left\lVert r^{k+1}\right\rVert_{2}$ $\leq$
$\sqrt{{2c_{2}\left\lVert{x}^{k+1}-{x}^{k}\right\rVert_{2}}/{s}}$ and
$\epsilon_{2}^{k}$ $\leq c_{1}\left\lVert\nabla g(x^{k+1})-\nabla
g(x^{k})\right\rVert_{2}$
$\leq c_{1}L\left\lVert x^{k+1}-x^{k}\right\rVert_{2}$ for $k\geq k_{0}$. By
the triangle inequality we have

(187)
$\begin{split}\left\lVert{x}^{k+1}-x^{\star}\right\rVert_{2}&\leq\bigg{\|}\text{prox}_{s_{k}h}(x^{k}-s_{k}\nabla
g(x^{k})-s_{k}\epsilon_{1}^{k})\\\
&\quad-\text{prox}_{s_{k_{0}}h}(x^{\star}-s_{k_{0}}\nabla
g(x^{\star})-s_{k_{0}}\epsilon_{1}^{k_{0}})\bigg{\|}_{2}+\left\lVert
r^{k+1}\right\rVert_{2}+\left\lVert r^{k_{0}+1}\right\rVert_{2}\\\
&\leq\bigg{\|}\text{prox}_{s_{k}h}(x^{k}-s_{k}\nabla
g(x^{k})-s_{k}\epsilon_{1}^{k})\\\
&\quad-\text{prox}_{s_{k_{0}}h}(x^{\star}-s_{k_{0}}\nabla
g(x^{\star})-s_{k_{0}}\epsilon_{1}^{k_{0}})\bigg{\|}_{2}+\left\lVert
r^{k+1}\right\rVert_{2}+\sqrt{\frac{2c_{2}\rho}{s}}\end{split}$

where we have used
$\left\lVert{x}^{k_{0}+1}-{x}^{k_{0}}\right\rVert_{2}\leq\rho$.

By the nonexpansivity of the proximal operator we have

(188)
$\begin{split}\left\lVert{x}^{k+1}-x^{\star}\right\rVert_{2}&\leq\bigg{\|}[x^{k}-s_{k}\nabla
g(x^{k})]-[x^{\star}-s_{k_{0}}\nabla g(x^{\star})]\bigg{\|}_{2}+\left\lVert
r^{k+1}\right\rVert_{2}+\sqrt{\frac{2c_{2}\rho}{s_{k_{0}}}}\\\
&\quad+s_{k}\left\lVert\epsilon_{1}^{k}\right\rVert_{2}+s_{k_{0}}\left\lVert\epsilon_{1}^{k_{0}}\right\rVert_{2}\\\
&\leq\bigg{\|}[x^{k}-s_{k}\nabla g(x^{k})]-[x^{\star}-s_{k_{0}}\nabla
g(x^{\star})]\bigg{\|}_{2}+\left\lVert
r^{k+1}\right\rVert_{2}+s_{k}\left\lVert\epsilon_{1}^{k}\right\rVert_{2}\\\
&\quad+\sqrt{\frac{2c_{2}\rho}{s_{k_{0}}}}+s_{k_{0}}c_{1}L\rho\end{split}$

By the nonexpansivity of the gradient descent operator, i.e.,
$\mathbf{I}-s\nabla g$, we obtain

(189) $\displaystyle\left\lVert{x}^{k+1}-x^{\star}\right\rVert_{2}$
$\displaystyle\leq\left\lVert x^{k}-x^{\star}\right\rVert_{2}+\left\lVert
r^{k+1}\right\rVert_{2}+s_{k}\left\lVert\epsilon_{1}^{k}\right\rVert_{2}+C_{\rho},\quad\forall
s_{k}\leq\frac{1}{L}$ (190) $\displaystyle=\left\lVert
x^{k}-x^{\star}\right\rVert_{2}+E^{k+1}+C_{\rho}.$

where $C_{\rho}=\sqrt{\frac{2c_{2}\rho}{s_{k_{0}}}}+s_{k_{0}}c_{1}L\rho$ and
$E^{k+1}=\left\lVert
r^{k+1}\right\rVert_{2}+s_{k}\left\lVert\epsilon_{1}^{k}\right\rVert_{2}$.
From (190) and by Definition 1, the sequence $\\{{x}^{k}\\}_{k\geq 1}$ is
quasi-Féjer relative to the set $X^{\star}$ if $\\{E^{k}\\}_{k\geq 1}$ is
positive and absolutely summable.

######  Theorem 10 (Quasi-Fejer monotonicity of the sequence generated by the
accelerated proximal gradient method).

Let $\\{x^{k}\\}_{k\geq 0}$ be the sequence generated by the approximate
accelerated proximal gradient (14) for solving problem (10) under Assumption 1
and with $s_{k}\leq\frac{1}{L}$. Assume we have summable iterative
displacements $\left\lVert x^{k}-x^{k-1}\right\rVert_{2}$ and that, for $k\geq
k_{0}$, we have $\epsilon_{2}^{k}\leq
c_{2}\left\lVert{x}^{k+1}-{x}^{k}\right\rVert_{2}\leq c_{2}\rho$ and
$\epsilon_{2}^{k}\leq c_{1}\left\lVert\nabla g(x^{k+1})-\nabla
g(x^{k})\right\rVert_{2}^{\top}$, then for any $x^{\star}\in X^{\star}$ and
$k\geq 0$ we have

(191) $\left\lVert{x}^{k+1}-x^{\star}\right\rVert_{2}\leq\left\lVert
x^{k}-x^{\star}\right\rVert_{2}+\left\lVert
x^{k}-x^{k-1}\right\rVert_{2}+E^{k+1}+C_{\rho}$

where $C_{\rho}=\sqrt{2Lc_{2}\rho}+c_{1}L\rho$. If $E^{k+1}:=\left\lVert
r^{k+1}\right\rVert_{2}+s_{k}\left\lVert\epsilon_{1}^{k}\right\rVert_{2}$ is a
positive and absolutely summable sequence, then $\\{x^{k}\\}_{k\geq 0}$ is a
quasi-Féjer sequence.

###### Proof.

For any optimal solution $x^{\star}$ of (10), we have

(192) $\displaystyle\left\lVert{x}^{k+1}-x^{\star}\right\rVert_{2}$
$\displaystyle=\left\lVert\text{prox}^{\epsilon_{2}^{k}}_{s_{k}h}(y^{k}-s_{k}\nabla^{\epsilon_{1}^{k}}g(y^{k}))-\text{prox}^{\epsilon_{2}^{k_{0}}}_{s_{k_{0}}h}(x^{\star}-s_{k_{0}}\nabla^{\epsilon_{1}^{k_{0}}}g(x^{\star}))\right\rVert_{2}.$

Rewriting $\text{prox}^{\epsilon_{2}^{k}}_{s_{k}h}(y)$ as
$\text{prox}_{s_{k}h}(y)+r^{k}$ and $\nabla^{\epsilon_{1}^{k}}g(y)$ as $\nabla
g(y)+\epsilon_{1}^{k}$ we obtain

(193)
$\begin{split}\left\lVert{x}^{k+1}-x^{\star}\right\rVert_{2}&=\bigg{\|}\text{prox}_{s_{k}h}(y^{k}-s_{k}\nabla
g(y^{k})-s_{k}\epsilon_{1}^{k})\\\
&\quad-\text{prox}_{s_{k_{0}}h}(x^{\star}-s_{k_{0}}\nabla
g(x^{\star})-s_{k_{0}}\epsilon_{1}^{k_{0}})+r^{k+1}-r^{k_{0}}\bigg{\|}_{2}.\end{split}$

By assumption we have $\epsilon_{2}^{k}\leq
c_{2}\left\lVert{x}^{k+1}-{x}^{k}\right\rVert_{2}$ and
$\epsilon_{2}^{k}\leq c_{1}\left\lVert\nabla g(x^{k+1})-\nabla
g(x^{k})\right\rVert_{2}\leq c_{1}L\left\lVert x^{k+1}-x^{k}\right\rVert_{2}$
for $k\geq k_{0}$. By the triangle inequality we have

(194)
$\begin{split}\left\lVert{x}^{k+1}-x^{\star}\right\rVert_{2}&\leq\bigg{\|}\text{prox}_{s_{k}h}(y^{k}-s_{k}\nabla
g(y^{k})-s_{k}\epsilon_{1}^{k})\\\
&\quad-\text{prox}_{s_{k_{0}}h}(x^{\star}-s_{k_{0}}\nabla
g(x^{\star})-s_{k_{0}}\epsilon_{1}^{k_{0}})\bigg{\|}_{2}+\left\lVert
r^{k+1}\right\rVert_{2}+\left\lVert r^{k_{0}+1}\right\rVert_{2}\\\
&\leq\bigg{\|}\text{prox}_{s_{k}h}(y^{k}-s_{k}\nabla
g(y^{k})-s_{k}\epsilon_{1}^{k})\\\
&\quad-\text{prox}_{s_{k_{0}}h}(x^{\star}-s_{k_{0}}\nabla
g(x^{\star})-s_{k_{0}}\epsilon_{1}^{k_{0}})\bigg{\|}_{2}+\left\lVert
r^{k+1}\right\rVert_{2}+\sqrt{\frac{2c_{2}\rho}{s}}\end{split}$

where we have used
$\left\lVert{x}^{k_{0}+1}-{x}^{k_{0}}\right\rVert_{2}\leq\rho$.

By the nonexpansivity of the proximal operator we have

(195)
$\begin{split}\left\lVert{x}^{k+1}-x^{\star}\right\rVert_{2}&\leq\left\lVert[y^{k}-s_{k}\nabla
g(y^{k})]-[x^{\star}-s_{k_{0}}\nabla g(x^{\star})]\right\rVert_{2}+\left\lVert
r^{k+1}\right\rVert_{2}\\\
&\quad+\sqrt{\frac{2c_{2}\rho}{s_{k_{0}}}}+s_{k}\left\lVert\epsilon_{1}^{k}\right\rVert_{2}+s_{k_{0}}\left\lVert\epsilon_{1}^{k_{0}}\right\rVert_{2}\\\
&\leq\left\lVert[y^{k}-s_{k}\nabla g(y^{k})]-[x^{\star}-s_{k_{0}}\nabla
g(x^{\star})]\right\rVert_{2}+\left\lVert r^{k+1}\right\rVert_{2}\\\
&\quad+s_{k}\left\lVert\epsilon_{1}^{k}\right\rVert_{2}+\sqrt{\frac{2c_{2}\rho}{s_{k_{0}}}}+s_{k_{0}}c_{1}L\rho\end{split}$

By the nonexpansivity of the gradient descent operator, i.e.,
$\mathbf{I}-s\nabla g$, we obtain

(196)
$\begin{split}\left\lVert{x}^{k+1}-x^{\star}\right\rVert_{2}&\leq\left\lVert
y^{k}-x^{\star}\right\rVert_{2}+\left\lVert
r^{k+1}\right\rVert_{2}+s_{k}\left\lVert\epsilon_{1}^{k}\right\rVert_{2}+C_{\rho,s_{k_{0}}},\quad\forall
s_{k}\leq\frac{1}{L}\\\ &=\left\lVert
x^{k}-x^{\star}+\beta_{k}(x^{k}-x^{k-1})\right\rVert_{2}+E^{k+1}+C_{\rho,s_{k_{0}}}\\\
&=\left\lVert x^{k}-x^{\star}\right\rVert_{2}+\left\lVert
x^{k}-x^{k-1}\right\rVert_{2}+E^{k+1}+C_{\rho,s_{k_{0}}},\end{split}$

where
$C_{\rho,s_{k_{0}}}=\sqrt{\frac{2c_{2}\rho}{s_{k_{0}}}}+s_{k_{0}}c_{1}L\rho$,
$E^{k+1}=\left\lVert
r^{k+1}\right\rVert_{2}+s_{k}\left\lVert\epsilon_{1}^{k}\right\rVert_{2}$ and
we used $\beta_{k}\leq 1$. From (196) and by Definition 1, the sequence
$\\{{x}^{k}\\}_{k\geq 1}$ is quasi-Féjer relative to the set $X^{\star}$ if
$\\{E^{k}\\}_{k\geq 1}$ is positive and absolutely summable provided we have
summable iterative displacements $\left\lVert x^{k}-x^{k-1}\right\rVert_{2}$.

## Acknowledgements

This work was supported by the Engineering and Physical Research Council
(EPSRC) grant EP/S000631/1 and the MOD University Defence Research
Collaboration (UDRC).

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$\displaystyle=2i\Big{[}-a^{(1)^{\prime}}_{0,+\omega}\left(a^{(1)}_{-1,+\omega}+a^{(1)}_{1,+\omega}\right)$
$\displaystyle+a^{(1)^{\prime}}_{0,-\omega}\left(a^{(1)}_{-1,-\omega}+a^{(1)}_{1,-\omega}\right)\Big{]}$
$\displaystyle+\frac{2i}{r}\cos$
$\displaystyle\big{(}\theta_{0}\big{)}\Big{[}a^{(1)}_{0,+\omega}\left(a^{(1)}_{1,+\omega}-a^{(1)}_{-1,+\omega}\right)$
$\displaystyle+a^{(1)}_{0,-\omega}\left(a^{(1)}_{-1,-\omega}-a^{(1)}_{1,-\omega}\right)\Big{]}$
$\displaystyle\lim_{\alpha\to
0}\langle\mathfrak{F}^{y}\rangle_{\chi,t}(\omega=\omega_{\text{br.}})$
$\displaystyle=0.$

This time the situation is reversed and
$\langle\mathfrak{F}^{x}\rangle_{\chi,t}$ vanishes. The resulting force is

$\displaystyle\lim_{\alpha\to
0}\mathbf{F}_{\text{trans}}(\omega=\omega_{\text{br.}})$
$\displaystyle=\alpha\mathbf{e}_{x}\int 2\pi
r\mathop{dr}\langle\mathfrak{F}^{x}\rangle_{\chi,t}.$ (89)

Importantly, now we have $\langle\mathfrak{F}^{x}\rangle_{\chi,t}\propto
1/\alpha$ in the limit $\alpha\to 0$ because of the $a^{(1)}_{0,\pm\omega}$
components. Hence,

$\lim_{\alpha\to
0}F_{\text{trans}}(\omega=\omega_{\text{br.}})\sim\alpha\cdot\frac{1}{\alpha}\sim\text{const.}$
(90)

Finally we consider the case $\omega=\omega_{\text{Kit.}}=b_{0}$. In this case
either $a^{(1)}_{-1,+\omega}$ and $a^{(1)}_{1,-\omega}$ or
$a^{(1)}_{1,+\omega}$ and $a^{(1)}_{-1,-\omega}$ will be purely imaginary and
$\propto 1/(i\alpha)$, see Eq. 39. If $\gamma<0$ and $b_{R}>0$,
$a^{(1)}_{-1,+\omega}$ and $a^{(1)}_{1,-\omega}$ are purely imaginary and we
have

$\displaystyle\lim_{\alpha\to
0}\langle\mathfrak{F}^{x}\rangle_{\chi,t}(\omega=\omega_{\text{Kit.}})$
$\displaystyle=2i\Big{(}a^{(1)^{\prime}}_{0,+\omega}a^{(1)}_{-1,+\omega}-a^{(1)^{\prime}}_{0,-\omega}a^{(1)}_{1,-\omega}\Big{)}$
$\displaystyle+\frac{2i}{r}\cos(\theta_{0})$
$\displaystyle\Big{(}a^{(1)}_{0,+\omega}a^{(1)}_{-1,+\omega}+a^{(1)}_{0,-\omega}a^{(1)}_{1,-\omega}\Big{)}$
$\displaystyle\lim_{\alpha\to
0}\langle\mathfrak{F}^{y}\rangle_{\chi,t}(\omega=\omega_{\text{Kit.}})$
$\displaystyle=0.$ (91)

The forces are therefore the same as in Eq. 89, but with
$\langle\mathfrak{F}^{x}\rangle_{\chi,t}$ as defined in App. K. Thus at the
Kittel resonance we also have

$\lim_{\alpha\to
0}F_{\text{trans}}(\omega=\omega_{\text{Kit.}})\sim\text{const.}$ (92)