Proof of Hölder's inequality in $l_p$

Question

I'm reading a proof of Hölder's inequality in $l_p = \{(a_k)\in\mathbb{R}^{\infty}:\sum a_k^p<\infty\}$ spaces.

Once it's homogeneous (that is, if it holds for $a = (a_k)$, $b = (b_k)\in l_p$ then it holds for $(\lambda a_k)$ and $(\mu b_k)$ for any $\lambda, \mu\in\mathbb{R}$), they assume $||a||_p=||b||_p = 1$ and go proving $\sum|a_kb_k|\le1$.

For this, they consider the graph of function $y=x^{p-1}\iff x=y^{q-1}$ and arbitrary $a_0,b_0\in \mathbb{R}_{\ge0}$ then calculate the areas $S_1=\int_0^{a_0}x^{p-1}=a_0^p/p$ and $S_2 = \int_0^{b_0}y^{q-1} = b_0^q/q$.

Then graphically they conclude $a_0b_0\le S_1+S_2$ (1), which gives us the inequality, by setting $a_0 = a_k, b_0 = b_k$ and summing it up $1$ to $n$ and taking limits.

I'm trying to prove inequality (1) algebrically, without the geometrical/graphical argument but my attempts are leading nowhere. Any hints?

Answer 1

Here is an elementary Calculus argument: let $f(x)=\frac {x^{p}} p +\frac {y^{q}} q-xy$ where $y=b_0$. Differentiate to check that $f$ is decreasing on $[0,y^{1/(p-1)})$ and then increasing. Thus $f$ attains its minimum at $x=y^{1/(p-1)}$. Some simple algebra will show that $f(y^{1/(p-1)})=0$ so the minimum value of $f$ is $0$. Hence $f(x) \geq 0$ fora ll $x \geq 0$. Put $x=a_0$.

Answer 2

(1) is the statement of Young's inequality. It follows from Jensen's inequality. Alternatively, you can replace following $\frac{1}{p}$ with $t$ and $\frac{1}{q}$ with $(1 - t)$ and use the concaveness of the logarithm function.