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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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    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.2012-01-26

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HINT:

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

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

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

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

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$2012-01-26