P-Norm of Real Sequence is Strictly Decreasing Function of P

Theorem

Let $p \ge 1$ be a real number.

Let ${\ell^p}_\R$ denote the real $p$-sequence space.

Let $\mathbf x = \sequence {x_n} \in {\ell^p}_\R$.

Suppose $\mathbf x$ is not a sequence of zero elements.

Let $\norm {\mathbf x}_p$ denote the $p$-norm of $\mathbf x$.

Then the mapping $p \to \norm {\mathbf x}_p$ is strictly decreasing with respect to $p$.

Proof

\(\ds \forall i \in \N: \, \)

\(\ds \sum_{n \mathop = 0}^\infty {\size {x_n} }\)

\(\ge\)

\(\ds \size {x_i}\)

Common Notion $5$: the whole is greater than the part

\(\ds \leadsto \ \ \)

\(\ds \forall i \in \N: \, \)

\(\ds \paren {\sum_{n \mathop = 0}^\infty {\size {x_n} }^p }^{\frac 1 p}\)

\(\ge\)

\(\ds \paren { {\size {x_i}^p } }^{\frac 1 p}\)

\(\ds \)

\(=\)

\(\ds \size {x_i}\)

\(\ds \leadsto \ \ \)

\(\ds \frac 1 p \map \ln {\sum_{n \mathop = 0}^\infty {\size {x_n} }^p}\)

\(\ge\)

\(\ds \map \ln {\size {x_i} }\)

\(\ds \leadsto \ \ \)

\(\ds \frac 1 p \map \ln {\sum_{n \mathop = 0}^\infty {\size {x_n} }^p} \sum_{i \mathop = 0}^\infty {\size {x_i} }^p\)

\(\ge\)

\(\ds \sum_{i \mathop = 0}^\infty {\size {x_i} }^p \map \ln {\size {x_i} }\)

Multiply both sides by $\size {x_i}$ and sum over $i \in \N$

By derivative of $p$-norm with respect to $p$:

\(\ds \dfrac \d {\d p} \norm {\mathbf x}_p\)

\(=\)

\(\ds \frac {\norm {\mathbf x}_p} p \paren {\frac {\sum_{n \mathop = 0}^\infty \size {x_n}^p \map \ln {\size {x_n} } } {\norm {\bf x}_p^p} - \map \ln {\norm {\bf x}_p} }\)

\(\ds \)

\(=\)

\(\ds \frac 1 {p \norm {\mathbf x}_p^{p \mathop - 1} } \paren {\sum_{n \mathop = 0}^\infty {\size {x_n} }^p \map \ln {\size {x_n} } - \sum_{n \mathop = 0}^{\infty} {\size {x_n} }^p \frac 1 p \sum_{i \mathop = 0}^\infty \map \ln {\size {x_i} } }\)

By $p$-Norm is Norm:

$\norm {\mathbf x}_p > 0$ for $\mathbf x \ne \sequence 0$.

By previously derived inequality, the term in parenthesis is negative.

This article, or a section of it, needs explaining. In particular: Label it and refer to it, and get that inequality into the same format so the reader hasn't got to do that. At the moment you need some mental gymnastics and it is far from obvious that they are the same thing (for as a start, the logarithm has moved from the outside to the inside of the summation, and that needs to be justified -- if it's even true. Can we even get the indices consistent, without having to switch between $i$ and $n$ and get confused as to which oen applies to which? Can we just call the index $k$ throughout? You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.

Hence:

$\forall p \ge 1: \dfrac \d {\d p} \norm {\mathbf x}_p < 0$

This article, or a section of it, needs explaining. In particular: Include a link to the appropriate result explaining why the result follows from the above You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.

$\blacksquare$