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 $\mathbf x$ where $p \ge 1$.

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

\(\ds \leadsto \ \ \)

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

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

\(\ge\)

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

\(\ds \)

\(=\)

\(\ds \size {x_i}\)

Equality holds only for a sequence of zero elements.

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

Then:

\(\ds \leadsto \ \ \)

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

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

\(>\)

\(\ds \size {x_i}\)

\(\ds \leadsto \ \ \)

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

\(>\)

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

\(\ds \leadsto \ \ \)

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

\(>\)

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

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

\(\ds \leadsto \ \ \)

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

\(<\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

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

\(<\)

\(\ds 0\)

Definition of Real P-Norm; Denote this inequality as $\paren \star$

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

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

\(=\)

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

\(\ds \)

\(=\)

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

By $\paren \star$, the term in parenthesis is negative.

By $p$-Norm is Norm:

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

Hence:

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

$\blacksquare$