I just went to check something about the $||\cdot||_\infty$ norm and realized that it can perhaps refer to two quite different things. I'm coming at this from an Analysis class so I am use to having $||f||_\infty = ess sup |f(t)|$. But, according to wikipedia (http://en.wikipedia.org/wiki/Supremum_norm) the uniform norm is the same notation but is defined as $||f||_\infty = sup |f(t)|$.

Unless I am really missing something, these are not equivalent. I'm a little uncomfortable with the possible ambiguity. How does one know which definition is actually being used in any one case? It seems like it is safe to say it is ess sup if anything explicitly references the $L^\infty$ space. I guesss I'm not sure what to ask. Can you verify that these are not necessarily equivalent? How do you differentiate which is being referred to?


1 Answer 1


Your feelings are correct.

The difference is in your underlying vector space. If its elements are functions (bounded, continuous, etc.), then it means $\sup$.

If its elements are equivalence classes (like in $L^\infty$), then you have to take ess-sup because there is a set of zero measure you have no control on.

  • $\begingroup$ Thanks that is helpful. This actually arose when I was going over Weierstrass's Theorem. In that case, it is sup right? And not ess sup? $\endgroup$
    – Fractal20
    Aug 7, 2013 at 16:29
  • 1
    $\begingroup$ Yes. In case of continuous functions, it is always sup. $\endgroup$
    – abatkai
    Aug 7, 2013 at 16:37
  • $\begingroup$ Actually, when it is about continuous functions in $L^\infty$, there is no other continuous function in its equivalence class. Hence, we are safe to think about the function even if it is its equivalence class. $\endgroup$
    – abatkai
    Aug 7, 2013 at 16:38

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