<p>First, we do need additional conditions on <span class="math-container">$f$</span> beyond the ones you supplied in the question. Consider <span class="math-container">$c(x)=\frac{1}{2}x^2$</span> and <span class="math-container">$f(x)=\tanh x$</span>, both restricted to the domain <span class="math-container">$\mathbb{R}_+$</span>. These functions satisfy your assumptions. However, there exists some <span class="math-container">$\bar{\alpha}$</span> large enough such that for all <span class="math-container">$\alpha&gt;\bar{\alpha}$</span>, <span class="math-container">$x^*(\alpha) =\text{argmax}_{x} \  \alpha f(x)-c(x)$</span> is a corner solution at <span class="math-container">$x^*(\alpha)=1$</span> so <span class="math-container">$x^*(\alpha)$</span> cannot be <span class="math-container">$\textit{strictly}$</span> increasing.</p>
<p>To get rid of corner solution issues, suppose that <span class="math-container">$$\text{Im}_{\alpha&gt;0} \ \text{argmax}_{x\in [0,1]}\{\alpha f(x)-c(x)\}\subseteq (0,1)$$</span> where <span class="math-container">$\text{Im}$</span> indicates the image of the argmax.</p>
<p>This condition requires further assumptions. Assume beyond your given assumptions that, <span class="math-container">$f''&lt;0$</span>, with <span class="math-container">$\lim_{x\to 1}f'(x)=0$</span> and <span class="math-container">$\lim_{x\to 0}f'(x)=\infty$</span>. The condition <span class="math-container">$\lim_{x\to 1}f'(x)=0$</span> contradicts your assumption that <span class="math-container">$f$</span> be strictly increasing over all of <span class="math-container">$\mathbb{R}_+$</span>. However, this is necessary for an interior solution over <span class="math-container">$[0,1]$</span> for all <span class="math-container">$\alpha &gt;0$</span>. We have a corner solution at <span class="math-container">$x=1$</span> if <span class="math-container">$\alpha f'(1)-c'(1)&gt;0 \implies \alpha &gt;\frac{c'(1)}{f'(1)}$</span>. If <span class="math-container">$f'(1)&gt;0$</span>, then there exists an <span class="math-container">$\tilde{\alpha}$</span> for which we get a corner solution and for all <span class="math-container">$\alpha&gt;\tilde{\alpha}$</span> we also get <span class="math-container">$x=1$</span> as a corner solution. This would mean the solution is not strictly increasing in <span class="math-container">$\alpha$</span> so we have to get rid of this behaviour.</p>
<p>Following Bertrand and BakerStreet's comments, assuming <span class="math-container">$f$</span> satisfies the above Inada conditions, any interior maximum satisfies the FOC <span class="math-container">$$0=\alpha f'(x^*(\alpha))-c'(x^*(\alpha)):=\mathscr{F}(x^*(\alpha),\alpha)$$</span> applying the implicit function theorem <span class="math-container">\begin{align*} \frac{dx^*(\alpha)}{d\alpha}=-\frac{\left(\frac{\partial \mathscr{F}}{\partial \alpha}\right)}{\left(\frac{\partial \mathscr{F}}{\partial x}\right)}=-\frac{f'(x^*(\alpha))}{\alpha f''(x^*(\alpha))-c''(x^*(\alpha))}&gt;0 \end{align*}</span></p>
<p>Where the inequality follows since <span class="math-container">$f',c''&gt;0$</span> and <span class="math-container">$f''&lt;0$</span>.</p>
<p>Therefore, assuming the solution is interior for all <span class="math-container">$\alpha&gt;0$</span>, strict concavity of <span class="math-container">$f$</span> is all you need.</p>

CC BY-SA 4.0
2024-05-06T18:11:03.327Z
Thanks! Assuming interiority it can also be shown that concavity is sufficient by rearranging the FOC to $$h(x):=\frac{c'(x)}{f'(x)}=\alpha$$ with $h$ strictly increasing and invertible, so the solution $x^*=h^{-1}(\alpha)$ is strictly increasing in $\alpha$, right?
raving-bandit
CC BY-SA 4.0
2024-05-06T19:28:08.550Z
@Joseph Basford Thank you for mentioning my comment!
BakerStreet
CC BY-SA 4.0
2024-05-07T06:40:52.673Z
@raving-bandit Yes, that is correct. That also makes the use of the Inada conditions explicit as they ensure $\text{Im}_{x\in(0,1)}(h)=\mathbb{R}_+$.
Joseph Basford
CC BY-SA 4.0
2024-05-06T09:56:05.277Z
2024-05-06T10:39:18.900Z
2024-05-06T10:39:18.900Z
Joseph Basford
<p>Let <span class="math-container">$f : R_+ \to [0,1]$</span> be continuously differentiable and strictly increasing with <span class="math-container">$f(0)=0, \lim_{x\to\infty}f(x)=1$</span> and let <span class="math-container">$c : R_+ \to R_+$</span> be continuously differentiable, strictly increasing, convex, and satisfy <span class="math-container">$c(0)=c'(0)=0$</span> and <span class="math-container">$\lim_{x \to \infty} c(x)=\lim_{x\to \infty}c'(x)=\infty$</span>.</p>
<p>What conditions on <span class="math-container">$f$</span> are needed so that the solution to</p>
<p><span class="math-container">$$ \max_{x\in[0,1]} \ \alpha f(x) - c(x) $$</span></p>
<p>is strictly increasing in <span class="math-container">$\alpha&gt;0$</span>? (This is certainly true if <span class="math-container">$f$</span> is linear.)</p>

CC BY-SA 4.0
2024-05-05T08:57:04.090Z
You could use the implicit function theorem to show this claim.
Bertrand
CC BY-SA 4.0
2024-05-05T16:07:30.137Z
You write $c : [0,1] \to R_+
$ and $c(1)=\infty$. There's a contradiction.
Michael Greinecker
CC BY-SA 4.0
2024-05-05T16:48:46.627Z
apologies, corrected.
raving-bandit
CC BY-SA 4.0
2024-05-05T17:28:04.057Z
@Bertrand I also had a hunch that this was the way to go, but I haven't found an obvious way to do it.
raving-bandit
CC BY-SA 4.0
2024-05-05T18:26:51.780Z
Hint: Consider the first order condition (if the second order conditions are met), this defines implicitly a function that gives argmax as $a$ varies, $x=x^*(a)$, using the implicit function theorem you can find the derivative $dx^*(a)/da$, and as a consequence the conditions for that derivative to be positive.
BakerStreet
CC BY-SA 4.0
2024-05-05T22:55:15.233Z
Still, a continuous function on a closed and bounded interval must be bounded.
Michael Greinecker
CC BY-SA 4.0
2024-05-06T03:17:53.583Z
@MichaelGreinecker you're correct again. Hopefully I've fixed it for good. Those conditions were meant to yield an interior solution, but the current set up should do too.
raving-bandit
<mathematical-economics><optimization>
When is argmax increasing in some multiplier of the objective function?
raving-bandit
# When is argmax increasing in some multiplier of the objective function?

(Asked by: username_0 on 2024-05-05)

Let $f : R\_+ \to [0,1]$ be continuously differentiable and strictly increasing with $f(0)=0, \lim\_{x\to\infty}f(x)=1$ and let $c : R\_+ \to R\_+$ be continuously differentiable, strictly increasing, convex, and satisfy $c(0)=c'(0)=0$ and $\lim\_{x \to \infty} c(x)=\lim\_{x\to \infty}c'(x)=\infty$.

What conditions on $f$ are needed so that the solution to

$$ \max\_{x\in[0,1]} \ \alpha f(x) - c(x) $$

is strictly increasing in $\alpha>0$? (This is certainly true if $f$ is linear.)

**Tags:** mathematical-economics, optimization
> **Comments:**
> **username_1**: You could use the implicit function theorem to show this claim.  
> **username_2**: You write $c : [0,1] \to R_+ $ and $c(1)=\infty$. There's a contradiction.  
> **username_0**: apologies, corrected.  
> **username_0**: @Bertrand I also had a hunch that this was the way to go, but I haven't found an obvious way to do it.  
> **username_3**: Hint: Consider the first order condition (if the second order conditions are met), this defines implicitly a function that gives argmax as $a$ varies, $x=x^*(a)$, using the implicit function theorem you can find the derivative $dx^*(a)/da$, and as a consequence the conditions for that derivative to be positive.  
> **username_2**: Still, a continuous function on a closed and bounded interval must be bounded.  
> **username_0**: @MichaelGreinecker you're correct again. Hopefully I've fixed it for good. Those conditions were meant to yield an interior solution, but the current set up should do too.  

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## 1 answers

### Answer by username_4 on 2024-05-06. Score: 1 | Accepted answer

First, we do need additional conditions on $f$ beyond the ones you supplied in the question. Consider $c(x)=\frac{1}{2}x^2$ and $f(x)=\tanh x$, both restricted to the domain $\mathbb{R}\_+$. These functions satisfy your assumptions. However, there exists some $\bar{\alpha}$ large enough such that for all $\alpha>\bar{\alpha}$, $x^\*(\alpha) =\text{argmax}\_{x} \ \alpha f(x)-c(x)$ is a corner solution at $x^\*(\alpha)=1$ so $x^\*(\alpha)$ cannot be $\textit{strictly}$ increasing.

To get rid of corner solution issues, suppose that $$\text{Im}\_{\alpha>0} \ \text{argmax}\_{x\in [0,1]}\{\alpha f(x)-c(x)\}\subseteq (0,1)$$ where $\text{Im}$ indicates the image of the argmax.

This condition requires further assumptions. Assume beyond your given assumptions that, $f''<0$, with $\lim\_{x\to 1}f'(x)=0$ and $\lim\_{x\to 0}f'(x)=\infty$. The condition $\lim\_{x\to 1}f'(x)=0$ contradicts your assumption that $f$ be strictly increasing over all of $\mathbb{R}\_+$. However, this is necessary for an interior solution over $[0,1]$ for all $\alpha >0$. We have a corner solution at $x=1$ if $\alpha f'(1)-c'(1)>0 \implies \alpha >\frac{c'(1)}{f'(1)}$. If $f'(1)>0$, then there exists an $\tilde{\alpha}$ for which we get a corner solution and for all $\alpha>\tilde{\alpha}$ we also get $x=1$ as a corner solution. This would mean the solution is not strictly increasing in $\alpha$ so we have to get rid of this behaviour.

Following Bertrand and BakerStreet's comments, assuming $f$ satisfies the above Inada conditions, any interior maximum satisfies the FOC $$0=\alpha f'(x^\*(\alpha))-c'(x^\*(\alpha)):=\mathscr{F}(x^\*(\alpha),\alpha)$$ applying the implicit function theorem \begin{align\*} \frac{dx^\*(\alpha)}{d\alpha}=-\frac{\left(\frac{\partial \mathscr{F}}{\partial \alpha}\right)}{\left(\frac{\partial \mathscr{F}}{\partial x}\right)}=-\frac{f'(x^\*(\alpha))}{\alpha f''(x^\*(\alpha))-c''(x^\*(\alpha))}>0 \end{align\*}

Where the inequality follows since $f',c''>0$ and $f''<0$.

Therefore, assuming the solution is interior for all $\alpha>0$, strict concavity of $f$ is all you need.

> **Comments:**
> **username_0**: Thanks! Assuming interiority it can also be shown that concavity is sufficient by rearranging the FOC to $$h(x):=\frac{c'(x)}{f'(x)}=\alpha$$ with $h$ strictly increasing and invertible, so the solution $x^*=h^{-1}(\alpha)$ is strictly increasing in $\alpha$, right?  
> **username_3**: @Joseph Basford Thank you for mentioning my comment!  
> **username_4**: @raving-bandit Yes, that is correct. That also makes the use of the Inada conditions explicit as they ensure $\text{Im}_{x\in(0,1)}(h)=\mathbb{R}_+$.  

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