2
$\begingroup$

Given that $f$ and $g$ are essentially bounded, show that their sum $f+g$ is also essentially bounded and furthermore show that the triangle inequality holds $||f+g||_{\infty}\leq ||f||_{\infty}+||g||_{\infty}$, where the norm in question is the $L_{\infty}$ norm. [The hint provided is that it is enough to show the inequality holds almost everywhere, which makes sense because equality in the $L_{\infty}$ norm can differ at most by a set of measure zero.]

I know that the definition of an essentially bounded function $f:E\to\overline{\mathbb{R}}$ is essentially bounded if $\exists A\in [0,\infty)$, where $|f|\leq A$ almost everywhere.

I know how to show the triangle inequlaity using Holder's inequality for $p-$norms, where $p<\infty$, but I'm not sure how to show it when $p=\infty$

$\endgroup$
  • $\begingroup$ The results for finite $p$ might actually be a distraction rather than help here. I would recommend first proving that, on the space of continuous functions $f$ (say, on $[0, 1]$), the supremum of $|f|$ (in fact, it turns out to be a max) is a norm on that space. To address $|| \cdot ||_{\infty}$ (i.e., to handle the behavior of a function on sets of small measure), you have to go from supremum to "essential supremum". math.stackexchange.com/questions/462006/… $\endgroup$ – avs Aug 8 '16 at 17:46
1
$\begingroup$

So $|f(x)|\le \|f\|_\infty$ for a.e. $x$, and $|g(x)|\le\|g\|_\infty$ for a.e. $x$. Let $A$ be the set where the first inequality holds and $B$ the set where the second inequality holds. Then the complement of $A\cap B$ has measure $0$. Try to estimate $|f(x)+g(x)|$ for $x\in A\cap B$.

$\endgroup$
0
$\begingroup$

It's not as complicated as Hoelder's inequality. Start by showing $\sup_\alpha{x_\alpha+y_\alpha}\leq \sup_\alpha x_\alpha + \sup_\alpha y_\alpha$ for any set of $x_\alpha$ and $y_\alpha$. Then incorporate the "almost everywhere" part.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.