The supremum norm is submultiplicative

2,022

Hint:$|f(x)g(x)|=|f(x)|\,|g(x)|\leq\|f\|_{\infty} |g(x)|,\forall x\in X\,\,\,$

Share:
2,022

Related videos on Youtube

htd
Author by

htd

Updated on August 01, 2022

Comments

  • htd
    htd over 1 year

    Is the following proof correct:

    Let $X$ be a compact Hausdorff space and $C(X)$ the space of continuous functions $f: X \to \mathbb{C}$. We can equip $C(X)$ with the (edit: sorry, semi-)norm $\lVert f \rVert_\infty = \sup_{x\in X} |f(x)|.$ I want to prove that for $f,g \in C(X)$, $\lVert fg \rVert_\infty \leq \lVert f \rVert_\infty \lVert g \rVert_\infty $. Let $z \in X$ be arbitrary then $$ \lvert f(z)g(z) \rvert \leq \lvert \; \sup_{x} \lvert f(x) \rvert g(z) \;\rvert=\sup_{x} \lvert f(x) \rvert \lvert g(z)\rvert \leq \sup_{x} \lvert f(x) \rvert \sup_{y} \lvert g(y) \rvert. $$

    Since $z$ was arbitrary $$ \sup_{z} f(z)g(z)\leq \sup_{x} f(x) \sup_{y} g(y) $$

    that is the sought.

    • Hamou
      Hamou about 9 years
      $\lVert f \rVert_\infty = \sup_{x\in X} f(x)$ is not a norm!
    • Igor Caetano Diniz
      Igor Caetano Diniz over 5 years
      Hamou? Is not a norm?
    • AJY
      AJY about 3 years
      @IgorCaetanoDiniz As written, it's not nonnegative, and will also give $0$ on functions which are nonzero (e.g. consider $f(x) = - x^2$ on some compact subset of $\mathbb{R}$ containing $0$). What you're looking for is $\sup_x |f(x)|$.
  • htd
    htd about 9 years
    That shortens it :) thanks.
  • Gaurang
    Gaurang over 2 years
    please could you elaborate to show, $||fg||$ $\le ||f|| \times ||g||$.