the continuity of argmin on convex function Define
$$x'=\text{argmin}_{x_1}f(x_1,\lambda),$$
where $f$ is a strictly convex function on $x_1$ and $\lambda$. I would like to ask if there is any theorem about the continuity of $x'$ w.r.t $\lambda$? If yes, can it be generalized to higher dimemions?
For example, I have
$$\mathbf{y}'=[y_1' ~~ y_2']^T=\text{argmin}_{y_1,y_2}g(y_1,y_2,\lambda),$$
where $g$ is a strictly convex function on $y_1$, $y_2$ and $\lambda$. Are $y_1'$ and $y_2'$ continuous w.r.t to $\lambda$?
 A: I think that you need to carefully state the conditions on the function $f$. There is not enough information, for instance, to ensure that the function is even defined: You need conditions ensuring that there is a minimum for each $\lambda$. Convex functions are very general creatures, and need not even be continuous.
However let us say that $f(x,\lambda):\mathbb{R}^{m+n}$ is 2-times continuously differentiable with respect to the variable $(x,\lambda)$ with $x\in\mathbb{R}^m$ and $\lambda\in\mathbb{R}^n$. 
Say that for $\lambda^1$ that there is a point $x^1$ such that $\nabla_x f(x^1,\lambda^1)=0$ and the Jacobian with respect to $x$ at this point $\frac{\partial f}{\partial x}$ is invertible. This is a minimum since $f$ is convex. Then there exists a neighbourhood $V$ around $\lambda^1$ such that we can we can define a continuously differentiable function g:
\begin{align}
g(\lambda)   &\triangleq x \text{ such that  $\nabla_x f(x,\lambda) = 0$} && \text{(implicit function theorem)}\\ 
             &=\text{argmin}_{x\in\mathbb{R}^m}f(x,\lambda) &&\text{(by convexity)}
\end{align}
This all follows from the implicit function theorem applied to the function $\nabla f(x,\lambda)$ and the equation $\nabla f(x,\lambda)=0$. I.e. yes, this argmin function is continuously differentiable under some circumstances.
The Jacobian $\partial_i \partial_j f$ needs to be invertible at this minimum $x_1$. Since $f$ is a convex function, the jacobian is symmetric nonnegative. The Jacobian will be invertible if it is positive definite: I.e. if $f$ is strongly convex in a convex neighbourhood of $x_1$ for $\lambda=\lambda_1$.
If you want to use weaker conditions, I'll see if I can find another theorem.
