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Minkowski's Inequality For Infinity

Can someone help with the proof of Minkowski inequality for $p=\infty$?

This is what I've done so far, but I seem confused as to what to do next.

So, I want to show that $$\|f+g\|_\infty \leq \|f \|_\infty + \|g\|_\infty.$$ By definition, $\|f\|_\infty=\inf\{M:|f(x)|\leq M~\text{a.e}\}$.

From the definition, I get that $|f|\leq M$ a.e and $|g|\leq N$ a.e. Thus $|f+g|\leq M+N$ a.e. I also know that $|f|\leq \|f\|_\infty$ a.e. and $|g| \leq \|g\|_\infty$ a.e. but I don't know how to bring everything together. I know I'm almost there, but I can't see the end unfortunately, so I need some help.

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marked as duplicate by t.b., Jonas Teuwen, Asaf Karagila, Henning Makholm, J. M. Nov 27 '11 at 4:01

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That certainly is not true. Take $f = 1$ and $g = -f$. –  Jonas Teuwen Nov 21 '11 at 9:31
    
sorry, I meant $\leq$ and not $=$. –  Joe Nov 21 '11 at 9:32
    
@Jonas surely your example gives 0 = ||1+(-1)|| <= ||1|| + ||(-1)|| = 0, so the inequality holds? –  Chris Taylor Nov 21 '11 at 9:34
    
what certainly is not true? –  Joe Nov 21 '11 at 9:34
    
@ChrisTaylor At first there was a $=$. –  Jonas Teuwen Nov 21 '11 at 9:59

1 Answer 1

up vote 3 down vote accepted

Let $A$ be the set where $|f(x)|$ is below $\|f\|_\infty$ and $|g(x)|$ below $\|g\|_\infty$. Then

$$|f(x) + g(x)| \leqslant |f(x)| + |g(x)| \leqslant \|f\|_\infty + \|g\|_\infty.$$

So this almost everywhere (only not on $\complement A$ which is of measure zero) hence we can take the supremum on $A$ to get

$$\|f + g\|_\infty \leqslant \|f\|_\infty + \|g\|_\infty.$$

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I don't get the conclusion. can you please explain. thanks –  Joe Nov 21 '11 at 9:43
    
@Joe Do you understand it now? I'm using the definition in the way that I only use what it means, no strict definition manipulation. –  Jonas Teuwen Nov 21 '11 at 10:06
    
Yes I do. Thanks. –  Joe Nov 21 '11 at 10:21

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