p-norms are ordered

In my linear analysis class p-norms were introduced in the context of normed vector spaces and at some point we had to show that the p-norms are ordered. This isn’t difficult to show using induction, but I’d like to emphasise a method I often use, which is to modify an expression into one that lends itself more easily to the tools from analysis.

A p-norm on a finite dimensional vector space over the field of non-negative reals is a mapping defined for p > 0 as follows:

\begin{aligned} || \cdot ||_p : \mathbb{R}_{+}^n \rightarrow \mathbb{R_{+}} \end{aligned}

\begin{aligned} ||x||_p = (\sum_{i=1}^{n} (x_i)^p)^{\frac{1}{p}} \end{aligned} 

Now, the key step in showing ||x||_p \geq ||x|| when 0<p  is to define

\begin{aligned} f: \mathbb{R}_{+}^n \rightarrow \mathbb{R_{+}} \end{aligned} 

\begin{aligned} f(\{x_i \}_{i=1}^{n}) = (\sum_{i=1}^{n} x_i)^p \end{aligned} 

\begin{aligned} g: \mathbb{R}_{+}^n \rightarrow \mathbb{R_{+}} \end{aligned} 

\begin{aligned} g(\{x_i \}_{i=1}^{n}) = \sum_{i=1}^{n} {x_i}^p \end{aligned} 

We note that f and g are smooth non-negative functions of severable variables that have global minima at \vec{0} \in \mathbb{R}_{+}^n . More importantly, we have:

\begin{aligned} \forall i, \frac{\partial f}{\partial x_i} = \frac{p}{(\sum_{i=1}^{n} x_i)^{1-p}} \end{aligned} 

\begin{aligned} \forall i, \frac{\partial g}{\partial x_i} = \frac{p}{(x_i)^{1-p}} \end{aligned} 

From this the general proof follows. I leave it to the reader to work out the details. If you’d like to check your work, my solution is available in the essays section.

I must note that the infinite dimensional case follows easily from the finite-dimensional case by considering divergent series, then convergent series.


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