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Using the fact that $$cd\leq\frac{c^p}{p}+\frac{d^q}{q}$$ if $$\frac{1}{p}+\frac{1}{q}=1$$ and letting $$c=\frac{|f(t)|}{\left(\int_{a}^{b}|f(t)|^p dt\right)^\frac{1}{p}}$$ and $$d=\frac{|g(t)|}{\left(\int_{a}^{b}|g(t)|^q dt\right)^\frac{1}{q}},$$ I have proven a lemma which states $$\int_{a}^{b}|f(t)g(t)|dt\leq\left(\int_{a}^{b}|f(t)|^p dt\right)^\frac{1}{p}\left(\int_{a}^{b}|g(t)|^q dt\right)^\frac{1}{q}.$$ But, how can this be used to prove the triangle inequality for this norm? I am a little confused about the $p$'s and $q$'s. Do I use what I know about the relationship of $p$ and $q$ to write the $q$'s as $p$'s?

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What happens if you apply your fact using the values of $c$ and $d$ you have, and integrate the resulting inequality from $a$ to $b$? – Aaron Jan 26 '12 at 4:47
I would post this as a comment if I could... This inequality is called Minkowski's inequality. You can find a proof using Holder's inequality (your lemma) on the wikipedia page for Minkowski's inequality. – zzz Jan 26 '12 at 5:38
perhaps this can interest you – leo Jan 26 '12 at 7:14
up vote 3 down vote accepted


From the lemma you proved (Hölder's Inequality). Let $f,g \in L_{p}[a,b]$.

Then $$\int_{a}^{b}|f+g|^{p}=\int_{a}^{b}|f+g|^{p-1}|f+g|$$

$$\le\int_{a}^{b}|f+g|^{p-1}(|f|+|g|) \text{ by triangle inequality of absolute value function}$$


From this step you can apply Hölder's inequality (for integrals), simplify by using the relationship of $p$ and $q$, then you will get your inequality called Minikwoski's Inequality.

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@Ashley, $p-1=p/q$ and $1-1/q=?\ldots$ – leo Jan 26 '12 at 7:09

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