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Let $f(x)$ be a concave function of many variables that we want to maximize $x=(x_1,x_2,\cdots,x_n)$ a point in its domain. Prove that if for some vector $y=(y_1,y_2,\cdots,y_n)$ we have that $f(x)\ge f(x+y)$ then for every number $t>1$, it holds that $f(x+y)\ge f(x+ty)$.

I know that $$f(\lambda x + (1 - \lambda)y)\ge \lambda f(x) + (1 - \lambda)f(y)\qquad \lambda\in[0,1]$$

I am having a really hard time, please help me. Thanks in advance.

Awnon Bhowmik
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Bill123
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1 Answers1

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Fix $t>1$.

Then in the inequality coming from the definition of $f$ being concave, replace $y$ with $x+ty$ and set $\lambda=\frac{t-1}{t}$

Brian Moehring
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  • I did what you said and it brings me to this: f (x + y) ≥ f (x) + f (y) – Bill123 Apr 26 '20 at 22:23
  • That is not even possible to get. Check your work again. If you want do it step by step and post your results for each step. – Brian Moehring Apr 26 '20 at 22:50
  • λx+(1−λ)y
    = λx+(1−λ)(x+ty) = λx+x+ty−λx-λty = x + ty - λty = x + ty - (t-1)ty/t = x + ty - (t-1)y = x + ty - ty + y = x + y     – Bill123 Apr 26 '20 at 23:00
  • Right, so this shows $f(\lambda x + (1-\lambda)y)$ in the inequality becomes $f(x+y)$. What does the expression on the other side of the inequality become? Any conclusions you can make from the whole thing? – Brian Moehring Apr 26 '20 at 23:09
  • the other end is λf(x) + (1-λ)f(x+ty), I don't know how to solve it if i set λ = (t-1)/t... – Bill123 Apr 26 '20 at 23:16
  • oooohhh now i get it!!! i just have to prove that f(x+ty) is concave too! Thanks man I really appreciate it! – Bill123 Apr 26 '20 at 23:21