Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For the infinity norm, for example in the definition of a regulated function on closed interval $A$. Where $\forall \epsilon,\, \, \exists\, \phi \, s.t. \|\phi - f\|_\infty < \epsilon$. The infinity norm is defined as $$\|\phi - f\|_\infty = \sup_{x \in A}|\phi(x) - f(x)|.$$ Can the two $x$'s be different? Or is this just the 'height' difference for any particular $x$?

share|cite|improve this question
"Can the two x's be different?" No. – Giuseppe Negro Apr 2 '12 at 13:39
The absolute value states the bind of the operator $\sup$. Also, $\sup_y|g(y)| + h(y)$ is a bad way to say $\sup_x|g(x)| + h(y)$. – AD. Apr 2 '12 at 13:52
Reminds me of a paper I once read where the author used the letter $m$ with different meanings all over. Something logically accepted, but not very pedagogical. – AD. Apr 2 '12 at 13:55
Well "x is the same x". Always use the same symbol in a proof, definition, theorem or etc. for the same object. Two object are equal if and only if they are the same! A question about this definition what do you assume to $\phi$, what is the regularity ? – checkmath Apr 2 '12 at 13:55
up vote 1 down vote accepted

No, the two $x$'s cannot be different. You are close in that $|\phi(x) - f(x)|$ is the height difference for any particular $x$. Then the $\sup$ says take the greatest of these.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.