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How would I prove the following theorem:

if $\sup \{ |f (x)| :x>0\} = 0$

this implies

$|f (x)| = 0$ for all real number $x$.

Adnan
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  • You need to edit your question. –  Mar 02 '15 at 08:01
  • @John what I need to prove is that if supremum of an absolute function over positive real line is equal to zero, this implies that the absolute of that function itself is zero over whole real line. – Adnan Mar 02 '15 at 08:09
  • I have just edited your question, is that what you are asking for? –  Mar 02 '15 at 08:32
  • Yes, I want to prove this. The extra information about f (x) is that it is continuous on positive real line. – Adnan Mar 02 '15 at 09:05

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What do you know about $f$? The statement is certainly not true for continuous functions on $\mathbb{R}$. A counter-example: $$ f(x) = |x| -x $$

MichalisN
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  • $f$ is continuous on $0$ to onward. – Adnan Mar 02 '15 at 08:56
  • In the case of function taken as $ f (x) = |x| - x $, for negative x , your assumption that $sup (f) = 0$ can be true in any case, so $sup (f) = 0$ condition is basically making your domain shrink. – Adnan Mar 02 '15 at 09:57
  • @user157336: Sorry I don't understand. The function $f$ is continuous on $\mathbb{R}$, $f(x)=0$ for all $x>0$ and $f(-1)=2$. How is this not a counter example to the statement you want to prove? – MichalisN Mar 02 '15 at 10:11
  • Yes indeed. The true statement is:

    if $\sup { |f (x)| :x>0} = 0$

    this implies

    $|f (x)| = 0$ for all real number $x > 0$.

     – Adnan Mar 02 '15 at 10:38