I want to show that $(L^\infty,\|\cdot\|_\infty)$ is a normed vector space. I understand that there are two things to show; firstly, that $L^\infty(X,\mu)$ is a linear space and secondly that $\|\cdot\|_\infty$ defines a norm on this space.

I don't have solutions to check whether what I have done here is correct or not - could somebody verify for me and point out anything I have done wrong, or that could be done better? In particular:

  1. In part (iii) of (b), how is it that taking out the factor of $|c|$ out of the $\inf$ doesn't affect the value of $M\in\mathbb R^+$ which bounds $|cf(x)|$?

  2. Similarly, in part (iv) of (b), should there be a $\le$" sign on the jump from line two to three when taking the $\inf$ of the sum?

(a) $L^\infty(X,\mu)$ is a linear space:

Take $f+g\in L^\infty(X,\mu)$ then $\exists\,M,N\ge 0:|f(x)|\le M\,,\forall x\in X\setminus A$ and $|g(x)|\le N\,,\forall x\in X\setminus B$, where $A,B$ are measurable sets of measure zero.

Consider then $|(f+g)(x)|\le|f(x)|+|g(x)|\le M+N,\,\forall x\in X\setminus(A\cup B)$. So, defining $P:=M+N\ge0$ we have shown that $|(f+g)(x)|\le P$. Thus, $f+g\in L^\infty(X,\mu)$.

Take then $f\in L^\infty(X,\mu)$ and some $c\in\mathbb R$. Then consider $|cf(x)|=|c||f(x)|\le|c|M=:Q$ where, clearly, $Q\ge0$. And so, it follows that $cf\in L^\infty(X,\mu)$.

Hence, $L^\infty(X,\mu)$ is closed under the addition and scalar multiplication of it's elements, and so is a linear space.

(b) $(L^\infty,\|\cdot\|_\infty)$ is a normed vector space:

We recall that $\|f\|_\infty=\inf\{M:|f(x)|\le M,\,\forall x\in X\setminus A\}$, where $f\in L^\infty(X,\mu)$.

(i) Suppose $f\in L^\infty(X,\mu)$ and consider $\|f\|_\infty=\inf\{M:|f(x)|\le M,\,\forall x\in X\setminus A\}$. By definition, such an $M$ is greater than or equal to zero, hence $\|f\|_\infty\ge0$, showing that our norm is nonnegative.

(ii) Consider: $\|f\|_\infty=0\iff\inf\{M:|f(x)|\le M,\,\forall x\in X\setminus A\}=0$

$\iff\forall x\in X\setminus A,\,|f(x)|\le0\iff\forall x\in X\setminus A,\,0\le f(x)\le0$ which means that $f\equiv0$ on $X\setminus A$. This establishes nondegeneracy of the norm.

(iii) For $c\in\mathbb R$ consider $\|cf\|_\infty=\inf\{M:|cf(x)|\le M,\,\forall x\in X\setminus A\}=\inf\{M:|c||f(x)|\le M,\,\forall x\in X\setminus A\}$

$=|c|\inf\{M:|f(x)|\le M,\,\forall x\in X\setminus A\}=|c|\|f\|_\infty$, showing multiplicativity of our norm.

(iv)Lastly, for $f,g\in L^\infty(X,\mu)$ consider,

$$\|f+g\|_\infty=\inf\{P:|f(x)+g(x)|\le P,\,\forall x\in X\setminus (A\cup B)\}$$

$$\le\inf\{M+N:|f(x)|+|g(x)|\le M+N,\,\forall x\in X\setminus (A\cup B)\}$$

$$=\inf\{M:|f(x)|\le M,\,\forall x\in X\setminus A\}+\inf\{N:|g(x)|\le N,\,\forall x\in X\setminus B\}$$

$$=\|f\|_\infty +\|g\|_\infty$$

So showing the triangle inequality.


I think you could be more careful with how you are treating the statements involving things that only hold on the complement of a measurable set.

E.g. I'd like better:

$\|f\|_\infty=\inf\{M:\exists A, \mu(A)=0 \colon \forall x\in X\setminus A\ |f(x)|\le M,\}$

Otherwise I might ask myself what the set $A$ is supposed to be. But this is just notation...

To (b) iii:

$\inf\{M\colon |c||f(x)|\leq M, ... \}=\inf\{M\colon |f(x)|\leq \frac{M}{|c|}, ... \}=\inf\{|c|\tilde{M}\colon |f(x)|\leq \tilde{M}, ... \}=|c|\cdot\inf\{\tilde{M}\colon |f(x)|\leq \tilde{M}, ... \}=\|f\|_\infty$

(at least for $c\neq 0$)

To (b) iv: yes I think you should use an $\leq$ sign

Otherwise I don't have concerns about your solutions, seems okay to me.

  • $\begingroup$ Thanks for the feedback. In the second line under (a) I mention that $A$ is a measurable set where $\mu(A)=0$; perhaps I should have noted it again in part (b) if that's what you mean(?). Also, $M$ isn't a set, it's a number which bounds from above the values of $|f(x)|$ when $x\in X\setminus A$, so I don't know what you mean when you write that you would rather see $\|f\|_\infty=\inf\{M:\exists M, \mu(M)=0 \colon \forall x\in X\setminus A\ |f(x)|\le M,\}$. Could you clarify, please? $\endgroup$ – Jeremy Jeffrey James Mar 30 '18 at 15:42
  • $\begingroup$ yeah with M you are totally right it was a typo I meant A I'll correct my answer $\endgroup$ – VanillaThunder Mar 30 '18 at 15:43
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    $\begingroup$ I understand perfectly what is going on with (b) part (iii) now; thanks for spelling that out for me! $\endgroup$ – Jeremy Jeffrey James Mar 30 '18 at 15:45
  • $\begingroup$ see the problem is (I think) it's not necessarily always the same set $A$, rather all these things hold for some set $A$ with Lebesgue measure zero, you're welcome $\endgroup$ – VanillaThunder Mar 30 '18 at 15:45
  • $\begingroup$ You're right that it's not necessarily the same set $A$ for different $f,g\in L^\infty(X,\mu)$. But given a particular $f\in L^\infty(X,\mu)$ you can say that there is some $A$ which has measure zero for which we have $|f(x)|\le M\,,\forall x\in X\setminus A$. $\endgroup$ – Jeremy Jeffrey James Mar 30 '18 at 15:49

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