Continuity of a parameterized convex optimization problem I have a parameterised optimization problem:
    \begin{align}
 \boldsymbol{S}(p)= &\arg \min_{ \boldsymbol{x}} g( \boldsymbol{x})\\
 \text{s.t. } &      \boldsymbol{A}(p) \textbf{x}  = \boldsymbol{b}(p)\\
 \end{align}
Where we have the following assumptions:
$ \boldsymbol{g}(x) $ is strongly convex, $ \boldsymbol{A}(p) $ is full row-rank.  $  \boldsymbol{A}(p) $ and $  \boldsymbol{b}(p) $ is $ C_1 $ continous.  
I now want to proove continuity of $\boldsymbol{S}(p)$.

I think I have a working solution strategy (I am not sure if it is precise enough yet)
1: Due to strict convexity, for a fixed p, a unique solution to the KKT conditions exists (which also is the minimum):
\begin{equation}\label{key}
\boldsymbol{R}( \boldsymbol{z}) = 
{\left[\begin{matrix}\nabla_{ \boldsymbol{x}}  \boldsymbol{g} (x) +   \boldsymbol{A}^T \boldsymbol{\lambda} \\   \boldsymbol{A} \textbf{x}  - \boldsymbol{b} 
 \end{matrix} \right]}=  \boldsymbol{0 }
\end{equation} 
For  $ \boldsymbol{z} ={\left[\begin{matrix}\boldsymbol{x }^T &  \boldsymbol{\lambda}^T \end{matrix} \right]}
  ^T $
Using the implicit function theorem to express $ \frac{\partial  \boldsymbol{z}}{\partial p} $ :
\begin{equation}
\frac{\partial  \boldsymbol{R} }{\partial  \boldsymbol{z} } \frac{\partial  \boldsymbol{z}}{\partial p} + \frac{\partial  \boldsymbol{R}}{\partial p} =0
\end{equation}
Thus:
\begin{equation}\label{SENSITIIVITY}
\frac{\partial  \boldsymbol{z}}{\partial p}  = - \left( \frac{\partial  \boldsymbol{R} }{\partial  \boldsymbol{z} } \right)^{-1}  \frac{\partial  \boldsymbol{R}}{\partial p}
\end{equation} '
The first term:
\begin{equation}
\frac{\partial  \boldsymbol{R} }{\partial  \boldsymbol{z} } = 
{\left[\begin{matrix}\nabla_x^2 ( g(x))  &  \boldsymbol{A}^T \\  \boldsymbol{A} &  \boldsymbol{0}  \end{matrix} \right]}
\end{equation}
Is invertible if g is strictly convex, and A full row-rank. 
Secondly  $ \frac{\partial  \boldsymbol{R}}{\partial p} $ exists if $ \boldsymbol{A}(p) $ and $ \boldsymbol{b}(p)$ is $C_1$. 
Which appearantly shows continuity of $  \boldsymbol{z} $ (and  $ \boldsymbol{x} $). 
I am unsure if this is precise enough or if I need extra conditions. Can anyone verify?.(or point out what I am missing if it does not hold) Any references or guidelines would be helpful.
Bonus question: By the above, then it seems that $\boldsymbol{z}$ is also $ C_1 $ continous, or do I need any new conditions to state this?.
 A: The underlying principle in obtaining continuity (even continuous differentiability) for $S$ with the IFT is correct. Personally, I would write things a bit differently/more precisely and in a different order and have done so below, but there is a gap: 
Before you talk about continuity of $S$, you should prove that $S$ is actually well defined, so that there indeed exists a (unique) solution to your optimization problem for every parameter $p$. Due to strict convexity of $g$ and linearity of the constraints w.r.t. $x$, it is clear that if a solution exists, then it is unique. But in fact, I think there are instances of your setup where there is no solution, such as $g(x_1,x_2) = e^{-x_1} + e^{-x_2}$, this is strictly convex, $A = (1,0)$ and $b = 1$.
If you can (under more assumptions, such as $A$ invertible, or properties of $g$ on the kernel of $A$, e.g. coercivity) prove that, then you can indeed proceed as you sketched. I think it should be done roughly as follows:


*

*The function $R$ should depend on $z$ and $p$. Then you note that the first component $x$ in $z$ is the unique solution to your optimization problem with parameter $p$, so $x = S(p)$, if and only if there is $\lambda$, the second component of $z$, such that $R(z,p) = 0$. This is due to (strong) convexity; the second component $\lambda$ in $z$ is then the associated Lagrange multiplier.

*Planning to use the implicit function theorem, you need to verify two things: that $R$ is Frechet-differentiable, and that the partial derivative $\partial R/\partial z$ is continuously invertible. (Look up the assumptions of the theorem!) You basically have already written down the argument for the latter.

*Now the implicit function theorem tells you that for every pair $(\bar z,\bar p)$ satisfying $R(\bar z,\bar p) = 0$, there is a continuously differentiable mapping $\varphi$ defined on some open set surrounding $\bar p$ such that $R(\varphi(p),p) = 0$ for all $p$ from that open set. Unwinding all the notation shows that in fact $S$ and $\varphi$ must coincide, so $S$ is also continuously differentiable. Note how the implicit function theorem requires to have a solution at hand first!

