1
$\begingroup$

In the following equation:$$f(y)=\sup_{x>0}\bigl(\exp(|y|-|y-x|)\bigr)$$ How can I find the value of supremum? can anyone help me to find it?
Thank you.

1 Answers 1

2

You want to maximize $|y| - |y-x|$ and since the contribution of $|y-x|$ is always $\leq 0$ while that of $|y|$ is always $> 0$ you see that the supremum is attained when $x = y$, and hence corresponds to $\exp(|y|)$.

EDIT: I haven't noticed the condition $x > 0$. As pointed out by Henry for $y < 0$ the supremum is attained when $x \rightarrow 0$ and is $\exp(0) = 1$.

  • 0
    @ blabler Thank you for your response, but why the supremum is in $x=y$?2012-11-30
  • 1
    Because this is when the contribution from the term $|y-x|$ is minimized2012-11-30
  • 0
    Strictly that is only true when $y \ge 0$. Otherwise the supremum is approached when $x$ approaches $0$, and so corresponds to $\exp(0)=1$.2012-11-30
  • 0
    @blabler Thanks for your answer!2012-11-30
  • 0
    @ Henry : Could you please explain more for $y<0$ ?2012-11-30
  • 0
    That's right, for $y < 0$ the supremum is 1, I haven't noticed the condition $x > 0$2012-11-30