(My problem is the same as problem 1.4 in this PDF.)
Some definitions:
Let $\rho_{\alpha,\beta}(\phi) = \sup_{x \in E_n} |x^{\alpha} D^{\beta} \phi(x) |$ where $\alpha, \beta$ are multi-indices and $E_n$ is an $n$-dimensional space. We index $\rho_{\alpha,\beta}$ by $\rho_k$ instead, and define $$ \rho = \sum_{k=1}^{\infty} 2^{-k} \frac{\rho_k}{1+\rho_k}. $$
I wish to show that $\rho$ is a seminorm. Therefore, I need to show it satisfies the three properties
- $\rho(\phi) \ge 0$; $\rho(\phi) = 0 \not \Leftrightarrow\phi = 0$.
- $\forall c \in \mathbb{R}, \quad \rho(c \phi) = |c| \rho(\phi)$.
- $\rho(\phi+\psi) \le \rho(\phi) + \rho(\psi)$
where $\phi, \psi \in S(E_n)$, the Schwartz class (a.k.a. testing functions).
I believe I've shown 1. and 2., except for the second property mentioned in 1., namely that $\rho= 0 \not\Rightarrow \phi = 0$.
However, my main question is to do with 3. Since $\rho_k$ is a norm, we have $\rho_k(\phi+\psi) \le \rho_k(\phi) + \rho_k(\psi)$. Hence,
\begin{align} \rho(\phi + \psi) & = \sum_{k=1}^{\infty} 2^{-k} \frac{\rho_k(\phi+\psi)}{1+\rho_k(\phi+\psi)} \\ &\le \sum_{k=1}^{\infty} 2^{-k} \frac{\rho_k(\phi)+\rho_k(\psi)}{1+\rho_k(\phi+\psi)} \\ & = \sum_{k=1}^{\infty} 2^{-k} \frac{\rho_k(\phi)}{1+\rho_k(\phi+\psi)} + \sum_{l=1}^{\infty} 2^{-l} \frac{\rho_l(\psi)}{1+\rho_l(\phi+\psi)}. \end{align}
I wish to show that this last line is less than or equal to the following expression:
$$ \sum_{k=1}^{\infty} 2^{-k} \frac{\rho_k(\phi)}{1+\rho_k(\phi)} + \sum_{l=1}^{\infty} 2^{-l} \frac{\rho_l(\psi)}{1+\rho_l(\psi)} $$
and the result would follow.
To do this, I would have to show that $\rho_k(\phi) \le \rho_k(\phi+\psi)$ for all $\phi,\psi$. So assume this is true (for the moment).
However, if we take $\psi = -\phi$, then
\begin{align} \rho_k(\phi+\psi) & = \rho_k(0) \\ & = 0. \end{align}
However, $\rho_k(\phi) \ge 0$ for all $\phi$. If our above statement were true, then $0 \le \rho_k(\phi) \le 0 \implies \rho_k(\phi) = 0 \iff \phi \equiv 0$. But this is a contradiction of our assumption. So the statement is not true for all $\phi, \psi$.
What is wrong with my counterexample? Is $\rho$ really a seminorm? How can I show this? This seminorm appears to be used consistently across different references.