P-Norm of Real Sequence is Strictly Decreasing Function of P
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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:
- $\forall p \ge 1 : \forall \mathbf x \ne \sequence 0 : \dfrac \d {\d p} \norm {\mathbf x}_p < 0$
$\blacksquare$