Deduction of Hölder Inequality

Question

If the Hölder inequality holds, we have $$ |x\cdot y|\leq \| x\|_p\| y\|_q $$ now if $y\neq 0$ this leads to $$ \frac{|x\cdot y|}{\| y\|_q}\leq \| x\|_p $$ Now my question. Is this implication true: $$ \sup_{y\in \mathbb{R}^n\backslash \{0\}}\frac{|x\cdot y|}{\| y\|_q}= \| x\|_p $$ If this implication is not true, is the identity at least true? If so, can you give a hint for a proof?

And besides that, is there any literature about this identity? I have searched a lot but I have found nothing. Thanks for any kind of help.

Answer 1

Your question is basically about when the Holder inequality attains equality. This is true if there exists a scalar $c$ such that $|y_i|^q = c|x_i|^p$ for all $i$.

So for example if $y_i = \text{sign}(x_i) \cdot x_i^{p/q}$ we have $|x \cdot y| = \sum_i |x_i|^{1+p/q} = \sum_i |x_i|^p$ and $\|y\|_q=\left(\sum_i |x_i|^p\right)^{1/q}$, whose ratio equals $\left(\sum_i |x_i|^p\right)^{1-1/q} = \left(\sum_i |x_i|^p\right)^{1/p} = \|x\|_p$.

Answer 2

According to the Hölder inequality $$ |x \cdot z| \leq\|x\|_{p}\|z\|_{q}, $$ from which follows $$ \|x\|_{p} \geq \frac{|x \cdot z|}{\|z\|_{q}} $$ and thus \begin{equation} \|x\|_{p} \geq \sup _{z \in \mathbb{R}^{n} \backslash\{0\}} \frac{|x \cdot z|}{\|z\|_{q}} . \tag{1} \end{equation} To prove $(1)$, it suffices to find a $z=z(x)$ that yields equality in $(1)$. For this we consider the following cases.

Case $p=1$. Let us set $$ z_{k}=\operatorname{sgn} x_{k}=\left\{\begin{array}{l} 1, x_{k} \geq 0 \\ -1, x_{k}<0 \end{array}\right. . $$ Then $\|z\|_{\infty}=1$ and. $$ x \cdot z=\sum_{k=1}^{n} x_{k} \operatorname{sgn} x_{k}=\sum_{k=1}^{n}\left|x_{k}\right|=\|x\|_{1}=\|z\|_{\infty}\|x\|_{1} . $$ Case $p=\infty$. There is an $i$ with $$ \left|x_{i}\right|=\operatorname{max}_{k}\left|x_{k}\right|=\|x\|_{\infty} $$ so we set $$ z_{i}=\operatorname{sgn} x_{i} \text { and } z_{k}=0 \text { for all } k \neq i . $$ Then $\|z\|_{1}=1$ and $$ x \cdot z=x_{i} \operatorname{sgn} x_{i}=\left|x_{i}\right|=\|x\|_{\infty}=\|z\|_{1}\|x\|_{\infty} \cdot $$ Case $p \in(1, \infty)$. Let's set $$ z_{k}=\left(\operatorname{sgn} x_{k}\right)\left|x_{k}\right|^{p-1}. $$ Then we have $$ x \cdot z=\sum_{k=1}^{n} x_{k} z_{k}=\sum_{k=1}^{n}(\operatorname{sgn} x_{k})\left|x_{k}\right|^{p-1} x_{k}=\sum_{k=1}^{n}\left|x_{k}\right|^{p}=\|x\|_{p}^{p} $$ and $$ \|z\|_{q}=\left(\sum_{k=1}^{n}\left|z_{k}\right|^{q}\right)^{1 / q}=\left(\sum_{k=1}^{n}\left|x_{k}\right|^{(p- 1) q}\right)^{1 / q}=\left(\sum_{k=1}^{n}\left|x_{k}\right|^{p}\right)^{1 / q}=\|x\|_{p}^{p / q}. $$ Thus $$ \|x\|_{p}\|z\|_{q}=\|x\|_{p}^{1+p / q}=\|x\|_{p}^{1+(p-1)}=\|x\|_{p}^{p}=|x \cdot z|, $$ which is what we wanted to show. $\quad \blacksquare$