We prove Holder's inequality: if $f \in L^p$ and $g \in L^q$ where $\frac{1}{p} + \frac{1}{q} = 1$ (with $1 < p, q < \infty$), then: $\int_X |fg| \, d\mu \leq \|f\|_p \|g\|_q$

Proof: We may assume $\|f\|_p = \|g\|_q = 1$ (otherwise normalize).

Step 1: We first establish Young's inequality: for $a, b \geq 0$, $ab \leq \frac{a^p}{p} + \frac{b^q}{q}$

This follows from the convexity of $e^x$. Let $a = e^s$ and $b = e^t$. Then: $e^{s+t} \leq \frac{e^{ps}}{p} + \frac{e^{qt}}{q}$

Setting $a^p = e^{ps}$ and $b^q = e^{qt}$ gives the result. Equality holds when $a^p = b^q$.

Step 2: Apply Young's inequality pointwise with $a = |f(x)|$ and $b = |g(x)|$: $|f(x)||g(x)| \leq \frac{|f(x)|^p}{p} + \frac{|g(x)|^q}{q}$

Step 3: Integrate both sides: $\int_X |f||g| \, d\mu \leq \frac{1}{p} \int_X |f|^p \, d\mu + \frac{1}{q} \int_X |g|^q \, d\mu$

Step 4: Since $\|f\|_p = \|g\|_q = 1$, the right side equals: $\frac{1}{p} + \frac{1}{q} = 1$

Thus: $\int_X |f||g| \, d\mu \leq 1 = \|f\|_p \|g\|_q$

Step 5: For general $f$ and $g$, apply the above to $f/\|f\|_p$ and $g/\|g\|_q$.