Uniform convergence of objective function implies convergence of minimizers Let $A,A_h\in M_{n\times m}(\mathbb{R})$ be $n\times m$ matrices with $\Vert (A-A_h)x\Vert\leq h^2 \Vert x\Vert$ for $h\in (0,1)$ and $e\in \mathbb{R}^n$.
Now consider for a fixed $\lambda>0$ the optimization problems
$$\min_xe^TA_{(h)}x+\lambda\Vert x\Vert_2^2\quad \text{subject to}\quad \Vert x\Vert_1\leq C_1, \Vert x\Vert_\infty\leq C_2.$$
I'll denote the minimizer of the Problem with $A_h$ by $x_h$ and of the one with $A$ by $x_0$. I already showed that $x_h$ converges to $x_0$. I would furthermore like to get some convergence rates, but so far I was only able to derive one for the case where $x_0$ is strictly feasible.
My approach was to use the KKT conditions, i. e
$$ 0\in A_{(h)}^Te+2\lambda x+\mu_1^{(h)}\partial\Vert x \Vert_1+\mu_2^{(h)}\partial\Vert x \Vert_\infty,$$
$$\mu_1^{(h)}, \mu_2^{(h)}\geq0,\quad \mu_1^{(h)}(\Vert x \Vert_1-C_1)=0,\quad \mu_2^{(h)}(\Vert x \Vert_\infty-C_2)=0$$ $$ \Vert x\Vert_1\leq C_1, \Vert x\Vert_\infty\leq C_2.$$
I calculated the subdifferentials:
$$\partial\Vert x \Vert_1=\{v:\Vert v\Vert_\infty\leq 1, v^Tx=\Vert x\Vert_1\}$$
$$\partial\Vert x \Vert_\infty=\{v:\Vert v\Vert_1\leq 1, v^Tx=\Vert x\Vert_\infty\}$$
We have $\Vert A^Te-A_h^Te\Vert\leq h^2 \Vert e\Vert$. Therefore when $x_0$ is strictly feasible, then because of the convergence I can assume that $x_h$ is strictly feasible as well. Hence $\mu_1^{(h)}=\mu_2^{(h)}=0$, so  I can easily derive $\Vert x_0-x_h\Vert\leq C h^2$.
Unfortunatly I can't come up with estimates in the other cases.
EDIT:
We can assume, that $x_0\geq 0$ (componentwise), otherwise we can transform the coordinate system. And we can restrict ourselves to the case, where $x_h$ lies in the same orthant as $x_0$, i.e $x_h\geq 0$.
For $x=(x_1,..,x_m)\geq 0 $ and $v\in \partial\Vert x\Vert_1$ we get $v_i=1$ when $x_i\not=0$ and $v_i\in [-1,1]$ else.
If we now look at the fairly simple case $\Vert x_0\Vert_1=\Vert x_h\Vert_1=C_1$ and $\Vert x_0\Vert_\infty=\Vert x_h\Vert_\infty<C_2$
with $x_0,x_h>0$, then we know that $\partial\Vert x_0 \Vert_1=\partial\Vert x_h \Vert_1=\{(1,..1)^T\}$. Hence
$$0= A_{(h)}^Te+2\lambda x+\mu_1^{(h)}(1,..1)^T.$$ But even in this simple case I don't know how to bound the convergence of $\mu_1^h\rightarrow \mu_1$.
EDIT:
I would furthermore like to add the condition $\sum_{i=1}^m x_i=0.$ Once again the same problem with the convergence of the multipliers arises. One can probably incooperate this condition into the problem, without adding a new condition. E.g. we can define the matrix $P$ by $x=P\tilde{x}$, where $\tilde{x}=(x_1,..,x_{m-1})^T$, and notice $\partial (f\circ P)(\tilde{x}) = P^T\partial f(P\tilde{x})$ if $f$ convex. I'll figure this out in more detail tomorrow!
EDIT:
I made some progress and could solve the case where $\Vert x\Vert_\infty \leq C_2$ is active. But I had to assume that $(x_1,..,x_{m-1})\geq 0$. I'm unsure if I can assume this w.l.o.g because of the condition $\sum_{i=1}^m x_i=0.$
I appreciate any advice.
 A: You can rewrite your objective function as
$$
e^TA_hx+\lambda\|x\|_2^2=\lambda\left\|\frac{v_h}{2\lambda}+x\,\right\|_2^2-\frac{v_h^Tv_h}{4\lambda}\ ,
$$
where $\ v_h=A_h^Te\ $, and your constraint set,
$$
\mathscr{C}=\big\{x\in\mathbb{R}^m\,\big|\,\|x\|_1\le C_1,\|x\|_\infty\le C_2\big\}\ ,
$$
is convex and closed. Your optimisation problem therefore reduces to finding the point $\ x\ $ in $\ \mathscr{C}\ $ which is closest (in Euclidean distance) to $\ -\frac{v_h}{2\lambda}\ $.  The optimiser, $\ x_h\ $ , must satisfy the inequality
$$
0\le\big(y-x_h\big)^T\left(\frac{v_h}{2\lambda}+x_h\right)
$$
for all $\ y\in\mathscr{C}\ $. In particular,
$$
0\le\big(x_0-x_h\big)^T\left(\frac{v_h}{2\lambda}+x_h\right)\ ,
$$
because $\ x_0\in\mathscr{C}\ $, and
$$
0\le\big(x_h-x_0\big)^T\left(\frac{v_0}{2\lambda}+x_0\right)\ ,
$$
because $\ x_h\in\mathscr{C}\ $. Adding these two inequalities gives
$$
0\le\frac{1}{2\lambda}\big(x_0-x_h\big)^T\big(v_h-v_0\big)+\big(x_0-x_h\big)^T\big(x_h-x_0\big)\ ,
$$
or
\begin{align}
\left\|x_h-x_0\right\|_2^2&\le\frac{1}{2\lambda}\big(x_0-x_h\big)^T\big(v_h-v_0\big)\\
&=\frac{1}{2\lambda}\big(x_0-x_h\big)^T\big(A_{h}-A\big)^Te\\
&=\frac{1}{2\lambda}e^T\big(A_{h}-A\big)\big(x_0-x_h\big)\\
&\le\frac{1}{2\lambda}\left\|e\right\|_2\left\|\big(A_{h}-A\big)\big(x_0-x_h\big)\right\|_2\ ,
\end{align}
by Cauchy-Schwarz, from which it follows that
\begin{align}
\left\|x_h-x_0\right\|_2^2
&\le\frac{h^2\|e\|_2\left\|x_h-x_0\right\|_2}{2\lambda}\ ,
\end{align}
and hence,
\begin{align}
\left\|x_h-x_0\right\|_2\le\frac{h^2\|e\|_2}{2\lambda}\ .
\end{align}
Note that the same argument works when $\ \mathscr{C}\ $ is any non-empty closed convex set, so adding the constraint $\ \sum_\limits{i=1}^mx_i=0\ $ will not cause any problems.
Derivation of inequality $\ 
0\le\big(y-x_h\big)^T\left(\frac{v_h}{2\lambda}+x_h\right)
$
To simplify the notation, put $\ v=\frac{v_h}{2\lambda}\ $. Then the optimisation problem is equivalent to minimising the function
\begin{align}
f(x)&=\|v+y\|^2\\
&=\|v+x_h +y-x_h\|^2\\
&=f(x_h)+2(y-x_h)^T(v+x_h)+\|y-x_h\|^2
\end{align}
over the constraint set. Since the constraint set is convex, then if $\ y\ $ is in it, so is $\ z_\alpha=\alpha y+(1-\alpha)x_h\ $ for any $\ \alpha\in[0,1]\ $. Therefore, $\ f(x_h)\le f(z_\alpha)\ $ for all such alpha.  Since $\ z_\alpha-x_h=\alpha(y-x_h)\ $, this is equivalent to
$$
0\le 2\alpha(y-x_h)^T(v+x_h)+\alpha^2\|y-x_h\|^2
$$
for all $\ \alpha\in[0,1]\ $, or
$$
-\frac{\alpha\|y-x_h\|^2}{2}\le (y-x_h)^T(v+x_h)
$$
for all $\ \alpha\in(0,1]\ $. The desired inequality now follows by taking the limit as $\ \alpha\rightarrow0^+\ $ on the left of the inequality immediately above.
