2

Could someone assist me in proving that the maximum of a convex function is also convex using the below definition? I cannot figure out how this proof would work and would appreciate some help!

$$ f(\lambda x +(1-\lambda)y) \le \lambda f(x)+(1-\lambda)f(y) $$

$$ \lambda \in[1,0] $$

Thank you!

mathforumpythonhelp
  • 31
  • 1
    You may find the answer here : https://math.stackexchange.com/questions/147475/proving-that-the-maximum-of-two-convex-functions-is-also-convex – David Reed Nov 20 '17 at 23:12

3 Answers3

1

Step 1: Take only two convex functions $f(x)$ and $g(x)$.

Step 2: Let $\lambda\in[0,1]$. Show that: \begin{align} \max\{ f(\lambda x +(1-\lambda) y),g(\lambda x + (1-\lambda )y)\} \leq \max\{ \lambda f(x) +(1-\lambda) f(y),\lambda g(x) + (1-\lambda)g(y)\} \end{align} Step 3: Show that $\max\{A+B,C+D\}\leq \max\{A,C\} + \max\{B,D\}$, for $A,B,C,D\in\mathbb{R}$

Step 4: Show that for $\alpha\geq 0$ we have: $\max\{\alpha A,\alpha B\}=\alpha\max\{A,B\}$, for $A,B\in\mathbb{R}$.

Step 5: Finish the proof. Repeat for more functions (induction).

Shashi
  • 8,738
  • For the induction, the base case is k=2, the induction hypothesis is that we have the result for k and the induction step is k+1. I'm not sure what to do when this appears in the induction step: Suppose $x^=\lambda x+(1-\lambda)y.$ Then $ f(x)\le max{\lambda f_1(x)+(1-\lambda)f(y),\dots ,f_{k+1}(x^)}$ because I think I can apply convexity on $f_{k+1}.$ What should I do? – user441848 Dec 31 '17 at 23:52
  • 1
    @Anneliset. I don't understand what you mean exactly with the last words. However what will help you much is to note the following $$\max{A, B, C} =\max{\max{A, B}, C} $$This makes the induction easy. – Shashi Jan 01 '18 at 00:25
  • There is a typo, I meant can't instead of can. – user441848 Jan 01 '18 at 02:02
  • @Anneliset. Okay, did the hint I gave you help? You can use it like this $$\max{f_1,...,f_{k+1}} =\max{\max{f_1,...,f_{k}},f_{k+1}}$$ you know that the maximum of two convex function is convex and you are done. – Shashi Jan 01 '18 at 09:59
  • 1
    Yes I think I got it, I'd have to associate the functions in pairs and then apply the base case which is that the maximum of 2 convex functions is convex. – user441848 Jan 01 '18 at 19:03
0

Let $f_{1}$ and $f_{2}$ be convex, and let $f=\max(f_{1},f_{2})$. $f(\lambda x+(1-\lambda)y)=\max({f_{1}(\lambda x+(1-\lambda)y),f_{2}(\lambda x+(1-\lambda)y)}) \leq \max(\lambda f_{1}(x)+(1-\lambda)f_{1}(y)); \lambda f_{2}(x)+(1-\lambda)f_{2}(y))=\lambda \max(f_{1}(x),f_{2}(x))+(1-\lambda) \max(f_{1}(y),f_{2}(y))=\lambda f(x)+(1-\lambda)f(y)$$

mich95
  • 8,713
0

I assume you have two convex functions $g$ and $h$ on some interval $I$, and you want to prove that $f$, defined as $f(x)=\max(g(x),h(x)), x\in I$ is convex too.

For $x,y \in I$, and $\lambda \in [0, 1] $, one of the functions $g$, $h$ is going to be the 'bigger' one at $(1-\lambda)x+\lambda y$, without loss of generality let's assume that it is $g$. Thus, $f((1-\lambda)x+\lambda y)=g((1-\lambda)x+\lambda y) \ge (1-\lambda)g(x)+\lambda g(y) \ge (1-\lambda)f(x)+\lambda f(y)$.