<p>As no one has answered, I will give this a try.</p>

<p>To begin with, together with compactness, and of course nonemptiness, of $X$, which you have assumed, you need $f(x,y)$ to be continuous in $x$ in order to ensure, via Weierstrass extrem value theorem, that $$\theta(y)=\text{argmin}_{x\in X}f(x,y)\tag{1}$$ exists for any $y$. It is good, as you do, to assume strict convexity in $x$, for this gives us a unique solutions to the minimization problems, and thus $(1)$ is a singleton. </p>

<p>Under these assumptions, I will try to show that $(1)$ is continous in the following sense, where I am essentially using <a href="https://en.wikipedia.org/wiki/Continuous_function#Definition_in_terms_of_limits_of_sequences" rel="nofollow noreferrer">the definition of continuity in terms of limits of sequences</a>. The fact that the argmin is a set makes it complicated to directly apply the notion of continuity of function which seldom, in elementary calculus, maps elements to some set of sets.</p>

<p><strong>Theorem.</strong> Consider a nonempty compact set $X\subseteq \mathbb{R}$ and a continuous function $f:\mathbb{R}^2\to\mathbb{R}$. Furthermore, suppose that $$\theta(\bar{y})=\{z|z\text{ minimizes }f(x,\bar{y})\text{ over }x\in X\}$$  is the singleton $\{\bar{z}\}$ (to ensure this, assume strict convexity in the first argument of $f$). Then, for sequences $z_n\in \theta(y_n)$, where $\{y_n\}_{\mathbb{N}}$ is a sequence converging to $\bar{y}$, we have $z_n\to \bar{z}$.</p>

<p><strong>Proof.</strong> I will give a short proof. It is by contradiction, and is a variant of the proof of Lemma 6.3.2. in Nonlinear Programming: Theory and Algorithms, 3rd Edition, 2006, by Bazaara et al. </p>

<p>So suppose $z_n\to\bar{z}$, $z_n\in \theta(y_n)$, and suppose $|z_n-\bar{z}|&gt;\epsilon&gt;0$ for all $n\in I$ where $I$ is some index set. Since $X$ is compact and $z_n\in X$, the sequence $\{z_n\}_I$ has a convergent subsequence $\{z_n\}_{I'}$ with limit $z_0$ in $X$. By assumption, $|z_0-\bar{z}|\geq\epsilon&gt;0$ and thus $z_0\neq \bar{z}$. In addition, for $\{y_n\}_{I'}$ it holds that $$f(z_n,y_n)\leq f(\bar{z},y_n)\tag{2}$$ since $z_n$ minimizes for $y_n$, but $\bar{z}$ does not minimize for $y_n$ necessarily. </p>

<p>Now, use continuity of $f$ in both of its arguments, and take the limit over $I'$ in $(2)$, to get that $$f(z_0,\bar{y})\leq f(\bar{z},\bar{y})\tag{3}.$$ But as $z_0\neq\bar{z}$ and $\bar{z}$ minimized $f(x,\bar{y})$, it follows that $z_0$ is another, distinct minimizer of $f(x,\bar{y})$ over $X$. Hence, $\theta(\bar{y})$ is not a singleton. But this contradicts the fact that $\theta(\bar{y})$ was a singleton. QED.</p>

<p><strong>Reference suggestion.</strong> <a href="https://en.wikipedia.org/wiki/Maximum_theorem" rel="nofollow noreferrer">The Theorem of the Maximum</a>. This theorem is more general and "provides conditions for the continuity of an optimized function and the set of its maximizers as a parameter changes". It involves the concept of a multivalued function (note that the assumption of the argmin being a singleton simplified the proof of the theorem above).</p>

CC BY-SA 3.0
2016-12-05T20:23:56.467Z
2017-02-11T10:17:15.630Z
2017-02-11T10:17:15.630Z
Elias
<p>Suppose for every $y$, $f(x,y)$ is strictly convex in $x$. $f(x,y)$ is continuous in $y$ and let $\mathcal X$ be compact (in my problem, $\mathcal X$ is an interval). Can anyone suggest any theorems or references for the problem of whether $x^*(y) = \text{argmin}_{x \in \mathcal X} f(x,y)$ is continuous in $y$?</p>

<macroeconomics><mathematical-economics>
Sensitivity/continuity of optimal solution
eulerup
# Sensitivity/continuity of optimal solution

(Asked by: username_0 on 2016-02-05)

Suppose for every $y$, $f(x,y)$ is strictly convex in $x$. $f(x,y)$ is continuous in $y$ and let $\mathcal X$ be compact (in my problem, $\mathcal X$ is an interval). Can anyone suggest any theorems or references for the problem of whether $x^\*(y) = \text{argmin}\_{x \in \mathcal X} f(x,y)$ is continuous in $y$?

**Tags:** macroeconomics, mathematical-economics

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## 1 answers

### Answer by username_1 on 2016-12-05. Score: 4 | Accepted answer

As no one has answered, I will give this a try.

To begin with, together with compactness, and of course nonemptiness, of $X$, which you have assumed, you need $f(x,y)$ to be continuous in $x$ in order to ensure, via Weierstrass extrem value theorem, that $$\theta(y)=\text{argmin}\_{x\in X}f(x,y)\tag{1}$$ exists for any $y$. It is good, as you do, to assume strict convexity in $x$, for this gives us a unique solutions to the minimization problems, and thus $(1)$ is a singleton.

Under these assumptions, I will try to show that $(1)$ is continous in the following sense, where I am essentially using [the definition of continuity in terms of limits of sequences](https://en.wikipedia.org/wiki/Continuous_function#Definition_in_terms_of_limits_of_sequences). The fact that the argmin is a set makes it complicated to directly apply the notion of continuity of function which seldom, in elementary calculus, maps elements to some set of sets.

**Theorem.** Consider a nonempty compact set $X\subseteq \mathbb{R}$ and a continuous function $f:\mathbb{R}^2\to\mathbb{R}$. Furthermore, suppose that $$\theta(\bar{y})=\{z|z\text{ minimizes }f(x,\bar{y})\text{ over }x\in X\}$$ is the singleton $\{\bar{z}\}$ (to ensure this, assume strict convexity in the first argument of $f$). Then, for sequences $z\_n\in \theta(y\_n)$, where $\{y\_n\}\_{\mathbb{N}}$ is a sequence converging to $\bar{y}$, we have $z\_n\to \bar{z}$.

**Proof.** I will give a short proof. It is by contradiction, and is a variant of the proof of Lemma 6.3.2. in Nonlinear Programming: Theory and Algorithms, 3rd Edition, 2006, by Bazaara et al.

So suppose $z\_n\to\bar{z}$, $z\_n\in \theta(y\_n)$, and suppose $|z\_n-\bar{z}|>\epsilon>0$ for all $n\in I$ where $I$ is some index set. Since $X$ is compact and $z\_n\in X$, the sequence $\{z\_n\}\_I$ has a convergent subsequence $\{z\_n\}\_{I'}$ with limit $z\_0$ in $X$. By assumption, $|z\_0-\bar{z}|\geq\epsilon>0$ and thus $z\_0\neq \bar{z}$. In addition, for $\{y\_n\}\_{I'}$ it holds that $$f(z\_n,y\_n)\leq f(\bar{z},y\_n)\tag{2}$$ since $z\_n$ minimizes for $y\_n$, but $\bar{z}$ does not minimize for $y\_n$ necessarily.

Now, use continuity of $f$ in both of its arguments, and take the limit over $I'$ in $(2)$, to get that $$f(z\_0,\bar{y})\leq f(\bar{z},\bar{y})\tag{3}.$$ But as $z\_0\neq\bar{z}$ and $\bar{z}$ minimized $f(x,\bar{y})$, it follows that $z\_0$ is another, distinct minimizer of $f(x,\bar{y})$ over $X$. Hence, $\theta(\bar{y})$ is not a singleton. But this contradicts the fact that $\theta(\bar{y})$ was a singleton. QED.

**Reference suggestion.** [The Theorem of the Maximum](https://en.wikipedia.org/wiki/Maximum_theorem). This theorem is more general and "provides conditions for the continuity of an optimized function and the set of its maximizers as a parameter changes". It involves the concept of a multivalued function (note that the assumption of the argmin being a singleton simplified the proof of the theorem above).


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