# 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 $\lambda$?

• This is clearly going to depend on the regularity of $f$, respectively $g$. A first rough result would be that if $f$ is $\lambda$-continuous in a an open set $U$, and it is strictly convex for any $\lambda \in U$, then $x'$ is $\lambda$-continuous in $U$; if it wasn't we would have a value of $\lambda$ for which $f$ has two different minimums.On the other hand if $f$ is not $\lambda$-continuous $x'$ doesn't have to be too. But I guess you are looking for something a bit more refined, right? Do you have any specific statement in mind? – Giovanni De Gaetano Feb 10 '15 at 9:27
• Chapter 7.E in Rockafellar & Wets, Variational Analysis deals with this. – Ronaldo Carpio Feb 11 '15 at 3:54
• @Giovanni De Gaetano yes. g(y_1,y_2,\lambda)=a(y_1,y_2)+\lambda b(y_1,y_2), where a is strictly convex and b is convex. Could you suggest any theorem about the continuity of the argmin to \lambda? – Timespace7 Feb 11 '15 at 14:13
• In your setting, without the additional assumption that $g(y_1,y_2,\lambda)$ is strictly convex, I don't even see why $\mathrm{argmin}$ should be a well defined function. In general it is defined to be the set of points where the argument achieves a minimum. If you could provide some more background perhaps I could be more helpful. And, as a side remark, if you enclose your latex code in dollar symbols $it is going to look at it's supposed to. – Giovanni De Gaetano Feb 11 '15 at 16:02 ## 1 Answer I think that you need to carefully state the conditions on the function$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^1such 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.

• Amazing answer...but can you find weaker conditions? I'd be very interested in this also... – AIM_BLB May 3 '17 at 14:45