The metric for $\mathcal {L^p}$ space is defined by: $$d(x,y)=\Big(\sum^{\infty}_{j=1} {\lvert \xi_{j}-\eta_j\rvert}^p\Big)^\frac{1}{p}$$ where, $x,y \ \in \ \mathcal {L^p}$ and $x=(\xi_j), y= (\eta_j)$.
Now, all of the axioms are readily proved by this metric provided that we prove that the R.H.S. of the metric is convergent.
The book that I am following (cited below), follows some steps, to prove the convergence. The first step is to derive an auxiliary inequality where we prove that for, some $\alpha$ and $\beta$ $\in \ \mathbb {R^+}$ and for some $p$ and $q$ which are conjugate exponents, the following inequality hold true, $$\alpha\beta \leq \frac{\alpha^p}{p} +\frac{\beta^q}{q}$$
Now, to prove this inequality the author assumes a function $u = t^{p-1}$ and with the help of this function, he proves the auxiliary inequality above.
$\bullet~$ My question is, why do we assume this particular function?
$\bullet~$ References:
- Introductory functional analysis with applications - Erwin Kreyszig Pg. 12-13
The inequality used in the context is Young's Inequality.
What you are speaking about is proving the inequality using a famous problem from Putnam and Beyond, namely
Let $f : [0, r) \rightarrow [0, \infty)$ be continuous and strictly increasing with $f(0) = 0$ (the case of $r = \infty$ is allowed). Show that for every $a$ in $[0, r)$ and every $b$ in image of $[0, r)$ under $f$, we have \begin{align*} ab \leqslant \int_{0}^{a} f(x) dx + \int_{0}^{b} f^{-1}(t) dt \end{align*}
In which you want to plug in the function $f(x) = x^{p - 1}$, which satisfies all the conditions to proof the Young's Inequality.
However I know an extension of the inequality which also gives us some important results.
The use of this inequality is a required one in the proof of $``l^p-\textit{norm being a norm}"$ through some steps and Lemmas.
What I mean is, if we have the linear space $(X, \| \cdot \|_{p})$, then $\| \cdot \|_{p}$ is a norm in $X$ (Where $X$ is a subspace of $\mathbb{K}^n$, for $\mathbb{K} = \mathbb{R} \backslash \mathbb{C}$).
$\bullet~$ Target: We will show using the Young's Inequality for the Hölder conjugate numbers $p, q$ $\in$ $(0, \infty)$, that $\| \cdot \|_{p}~$ is a norm.
$\bullet~$ Prerequisites and Defn : Let's consider the numbers $p, q$ $\in$ $(0, \infty)$ such that the following is satisfied. $$ \frac{1}{p} + \frac{1}{q} = 1 $$ Then assume the Young's Inequality for some $a, b$ $\in$ $\mathbb{K},~$ with $a, b \geqslant {0}$. $$ ab \leqslant \frac{a^p}{p} + \frac{b^q}{q} $$ Take any $x = (x_{1}, x_{2}, \dots, x_{n}) \in \mathbb{K}^n $. Define $\| x\|_{p}$ and $\| x\|_{q}$ in the usual manner, i.e., $$ \| x\|_{p} = \sqrt[p]{\sum_{k = 1}^{n} \lvert x_{i}\rvert^{p}} $$
$\bullet~$ Lemma 1: Consider $\boldsymbol{a}, \boldsymbol{b}$ $\in$ $\mathbb{K}^n$. Suppose $\boldsymbol{a} = (a_{1}, a_{2}, \dots, a_{n})$ and $\boldsymbol{b} = (b_{1}, b_{2}, \dots, b_{n})$. Then the following holds $$ \sum_{k = 1}^{n} \lvert a_{k} b_{k} \rvert \leqslant \| \boldsymbol{a}\|_{p} \|\boldsymbol{b} \|_{q} $$
$\bullet~$ Proof:
$\circ$ Case (1): The case with $\boldsymbol{a} = \boldsymbol{b} = \boldsymbol{0}$ is trivial.
$\circ$ Case (2): Let's consider $\| \boldsymbol{a}\|_{p} = \| \boldsymbol{b}\|_{q} = 1$. Then by Young's Inequality we have that \begin{align*} \lvert a_{i}b_{i} \rvert &\leqslant \frac{\lvert a_{i} \rvert^{p}}{p} + \frac{\lvert b_{i} \rvert^{q}}{q} \\ \implies \sum_{i = 1}^{n} \lvert a_{i} b_{i} \rvert &\leqslant \sum_{i = 1}^{n} \frac{\lvert a_{i} \rvert^{p}}{p} + \sum_{i = 1}^{n} \frac{\lvert b_{i} \rvert^{q}}{q} = \frac{1}{p} + \frac{1}{q} = 1 = \| \boldsymbol{a} \|_{p} \| \boldsymbol{b} \|_{q} \end{align*}
$\circ$ Case (3): Let's assume neither of $~\boldsymbol{a}, \boldsymbol{b}$ is $\boldsymbol{0},~$ nor $~\|\boldsymbol{a} \|_{p}, \|\boldsymbol{b} \|_{q} = 1 $.
Then let's pick $~\boldsymbol{x} = \dfrac{\boldsymbol{a}}{\|\boldsymbol{a} \|_{p}}~$ and $~\boldsymbol{y} = \dfrac{\boldsymbol{b} }{\|\boldsymbol{b} \|_{q}}$. Note that $\| \boldsymbol{x} \|_{p} = \| \boldsymbol{y} \|_{q} = 1.~$ So by the argument in Case (2), we have $$ \sum_{i = 1}^{n} \lvert x_{i} y_{i} \rvert \leqslant 1 \implies \sum_{i = 1}^{n} \frac{\lvert a_{i}b_{i} \rvert }{\| \boldsymbol{a}\|_{p} \cdot \|\boldsymbol{b} \|_{q}} \leqslant 1 \implies \sum_{i = 1}^{n} \lvert a_{i}b_{i} \rvert \leqslant \| \boldsymbol{a} \|_{p} \cdot \| \boldsymbol{b} \|_{q} $$
$\bullet~$ Lemma 2: Consider $\boldsymbol{a}, \boldsymbol{b} \in \mathbb{K}^n$ like before. then the following holds
$\bullet~$ (a) $$\|\boldsymbol{a + b} \|_{p}^{p} \leqslant \big( \|\boldsymbol{a} \|_{p} + \|\boldsymbol{b} \|_{p} \big)\cdot \|\boldsymbol{v} \|_{q} $$ Where $\boldsymbol{v} = \big((a_{1} + b_{1})^{p -1}, (a_{2} + b_{2})^{p - 1}, \dots, (a_{n} + b_{n})^{p - 1} \big)$
$\bullet~$ (b) The following inequality holds $$ \|\boldsymbol{a} + \boldsymbol{b} \|_{p} \leqslant \|\boldsymbol{a} \|_{p} + \|\boldsymbol{b} \|_{p} $$
$\bullet~$ Proof (a): Let's consider $(a_{i} + b_{i})^{p - 1} = v_{i}$ and $\boldsymbol{v} = (v_{1}, v_{2}, v_{3}, \dots, v_{n})$.
Then from Lemma 1 we have that \begin{align*} \| \boldsymbol{a} + \boldsymbol{b} \|_{p}^{p} =&~ \sum_{i = 1}^{n} \lvert (a_{i} + b_{i})^p \rvert \\ =&~\sum_{i = 1}^{n}\lvert (a_{i} + b_{i}) v_{i} \rvert \\ \leqslant&~ \sum_{i = 1}^{n} \lvert a_{i} v_{i} \rvert + \sum_{k = 1}^{n} \lvert b_{i}v_{i} \rvert\\ \leqslant&~ \|\boldsymbol{a} \|_{p} \cdot \|\boldsymbol{v} \|_{q} + \|\boldsymbol{b} \|_{p} \cdot \|\boldsymbol{v} \|_{q} \\ =&~ (\|\boldsymbol{a} \|_{p} + \|\boldsymbol{b} \|_{p}) \cdot \|\boldsymbol{v} \|_{q} \end{align*} Completing the proof.
$\bullet~$ Proof (b): Let $\boldsymbol{v}$ be the same as Lemma 2 (a), then from the relation of Hölder's Conjugate Numbers we have
$$ \frac{1}{p} + \frac{1}{q} = 1 \implies q \cdot (p - 1) = p $$ Again we have that $$ (pq - q) = p \implies \bigg(p - \frac{p}{q}\bigg) = 1 $$ Now we have that $(a_{i} + b_{i})^{p - 1} = v_{i}.~$ Therefore we have that $$ \| \boldsymbol{v}\|_{q}^{q} = \sum_{i = 1}^{n} \lvert (a_{i} + b_{i})^{(p - 1)\cdot q} \rvert = \|\boldsymbol{a + b} \|_{p}^{p} $$ Again, from Lemma 2 (a) we have that $$ \|\boldsymbol{a} + \boldsymbol{b} \|_{p}^{p} \leqslant \big( \| \boldsymbol{a}\|_{p} + \|\boldsymbol{b} \|_{p} \big) \cdot \| \boldsymbol{a + b} \|_{p}^{\frac{p}{q}} $$ If $\|\boldsymbol{a + b} \|_{p} = 0 $, then it's trivial. So let's assume that $\|\boldsymbol{a + b} \|_{p} > 0$. Then we have that $$ \| \boldsymbol{a + b} \|_{p}^{p - \frac{p}{q}} \leqslant \big( \| \boldsymbol{a}\|_{p} + \|\boldsymbol{b} \|_{p} \big) \implies \| \boldsymbol{a + b} \|_{p} \leqslant \big( \| \boldsymbol{a}\|_{p} + \|\boldsymbol{b} \|_{p} \big) \quad \bigg[\text{as } p - \frac{p}{q} = 1\bigg] $$
Hence we have proved Lemma 2 (a) & (b).
$\blacksquare~$ With these we have that for $\boldsymbol{a} = (a_{1}, a_{2}, \dots, a_{n}) \in \mathbb{K}^n$
$\bullet~$ $\| \boldsymbol{a}\|_{p} \geqslant 0$ with equality when $\boldsymbol{a} = \boldsymbol{0}$.
$\bullet~$ For any $\lambda$ $\in \mathbb{K}$, we have $$ \| \lambda \boldsymbol{a} \|_{p} = \sqrt[p]{\sum_{k = 1}^{n} \lvert \lambda a_{i}\rvert^{p}} = \lvert \lambda \rvert \sqrt[p]{\sum_{k = 1}^{n} \lvert a_{i}\rvert^{p}} = \lvert \lambda \rvert \|\boldsymbol{a} \|_{p} $$
$\bullet~$ The Triangle Inequality holds for any $\boldsymbol{b} = (b_{1}, b_{2},\dots, b_{n}) \in \mathbb{K}^n$, i.e., $$ \|\boldsymbol{a} + \boldsymbol{b} \|_{p} \leqslant \|\boldsymbol{a} \|_{p} + \|\boldsymbol{b} \|_{p} $$
$ \bullet~ \bullet~ \bullet ~$ Hence The map $\| \cdot \|_{p} : \mathbb{K}^n \to \mathbb{R}_{0}$ is a norm on $\mathbb{K}^n$.