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

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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 $p$.


\(\ds \forall i \in \N: \, \) \(\ds \sum_{n \mathop = 0}^\infty {\size {x_n} }\) \(\ge\) \(\ds \size {x_i}\) 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.


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