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?
$\|f+g\|_p\leq\|f\|_p+\|g\|_p$ where $\|f\|_p=(\int_{a}^{b}|f(t)|^p dt)^\frac{1}{p}$
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real-analysis
norm
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0What happens if you apply your fact using the values of $c$ and $d$ you have, and integrate the resulting inequality from $a$ to $b$? – 2012-01-26
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0perhaps [this](http://math.stackexchange.com/q/87636/8271) can interest you – 2012-01-26
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2I 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
1 Answers
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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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1@Ashley, $p-1=p/q$ and $1-1/q=?\ldots$ – 2012-01-26