Proof of Hölder's Inequality
Theorem: Let \(1\le p\le \infty\) and let \(p^*\) be the exponential conjugate of p. Let E be a measurable set and \(f\in L^p(E)\) and \(g\in L^{p^*}(E)\). Then
$$ \| fg \|_1 \le \|f\|_p\|g\|_{p^*} $$
which could also be written
$$ \int_E |fg| \le \left(\int_E |f|^p\right)^{1/p}\left(\int_E |g|^{p^*}\right)^{1/p^*} $$
Equality holds if and only if there are scalars \(\alpha,\beta\) such that \(\alpha |f|^p = \beta|g|^{p^*}\) a.e.
Proof (from Young's inequality): Young's inequality tells us that under these conditions
$$ |fg| \le \frac{|f|^p}{p} + \frac{|g|^{p^*}}{p^*} $$
We will initially assume that \(\|f\|_p=1=\|g\|_{p^*}\). Once we have this result, we will use the good ol’ trick of taking a more general "vector" and normalize it.
Now if \(1<p<\infty\) then both p and q are finite then the result follows by integrating this inequality. (Hint: after integration use the assumption of unit norm. If the result you get is 1, then you are done (why?).)
On the other hand if p=1 then \(q=\infty\). But in that case the proof is even easier.
$$ \int_E|fg|\le\int_E |f|\|g\|_\infty = \|g\|_\infty\int_E|f| = \|f\|_1\|g\|_\infty $$
This is clearly symmetric so that if \(p=\infty\) then q=1, so that the same idea of the proof applies here.
So we now have proved the theorem when the norm of each function is 1. We move on to consider functions of any norm. In that case \( \frac{f}{\|f\|_p} \) has norm 1 and so does \(\frac{g}{\|g\|_{p^*}}\). Therefore
$$ \int_E \left| \frac{fg}{\|f\|_p\|g\|_{p^*}}\right| \le \left\| \frac{f}{\|f\|_p} \right\|_p \left\| \frac{g}{\|g\|_{p^*}} \right\| $$
On both sides of this we can use linearity to remove the norms in the denominators, and therefore they cancel on each side, resulting in the desired inequality.
The above demonstrates the inequality, so we proceed to show the final statement, which is that
$$ \int_E |fg| = \left( \int_E|f|^p \right)^{1/p} \left(\int_E |g|^{p^*} \right)^{1/p^*} $$
holds if and only if there are scalars \(\alpha,\beta\) such that \(\alpha|f|^p=\beta|g|^{p^*}\). The \(\Leftarrow\) direction is very easy to show merely by plugging into the integral equality. So we only show the \(\Rightarrow\) direction, and to that end we now assume that \( \int_E |fg| = \left( \int_E|f|^p \right)^{1/p} \left(\int_E |g|^{p^*} \right)^{1/p^*} \).
It is essential that we know, for nonnegative functions \(h,k\) such that \(h\le k\) a.e., that if \(\int_E h = \int_E k\) then h=k a.e. We will soon apply this principle with \( h=\frac{|f|^p}{p} \) and \(k=\frac{|g|^{p^*}}{p^*} \).
We again assume that \(\|f\|_p=1=\|g\|_{p^*}\) and so our assumption amounts to
$$ \int_E |fg| = 1 = \frac{\int_E|f|^p}{p} + \frac{\int_E|g|^{p^*}}{p^*} $$
Now using the knowledge that \(|fg|\le \frac{|f|^p}{p}+\frac{|g|^{p^*}}{p^*}\) and the principle we articulated just above, we have
$$ |fg| = \frac{|f|^p}{p} + \frac{|g|^{p^*}}{p^*} $$
From here, using the equality condition from Young's inequality, we must have that there are scalars \(\alpha,\beta\) such that \(\alpha|f|^p = \beta|g|^{p^*}\) a.e.
As before, since we have shown this when the two functions have norm equal to 1, it is easy to take a more generalized vector and apply this result to its normalization, which quickly delivers the more generalized result.