Let $E$ be a topological vector space and $\{p_1,p_2,\dots\}$ be a nondecreasing family of seminorms on $E$. Define $$d(x,y)=\sum_{i=1}^{\infty} a_i \frac{p_i(x-y)}{1+p_i(x-y)},$$ where $\sum_{i=1}^{\infty} a_i<\infty$.
My question: How to prove that $p(x)=d(x,0)$ is not a seminorm?
It's easy to see that $p(x+y)\leq p(x)+p(y)$. Thus, the property that must fail is homogeneity, that is, $p(\lambda x) = |\lambda| p(x)$ for all $\lambda \in \mathbb{C}$ and all $x \in E$. We have, $$p(\lambda x)= \sum_{i=1}^{\infty} a_i \frac{p_i(\lambda x)}{1+p_i(\lambda x)}=\sum_{i=1}^{\infty} a_i \frac{|\lambda|p_i(x)}{1+|\lambda|p_i( x)}.$$
What algebraic manipulation could I do to get $p(\lambda x)\neq |\lambda|p(x)|$?
Choose any $x$ such that there exists $i \in \mathbb{N}$ such that $p_i(x) > 0$. Then for $|\lambda| > 1$, $$p(\lambda x) = \sum_{j=1}^{\infty} a_j \frac{|\lambda|p_j(x)}{1+|\lambda|p_j( x)} < \sum_{j=1}^\infty \frac{a_j |\lambda|p_j(x)}{1+p_j(x)} = |\lambda|p(x)$$ since $1+|\lambda|p_j(x) \geq 1 + p_j(x)$ for all $j$ and $1+|\lambda|p_i(x) > 1 + p_i(x)$.