Rigourously proving continuity of multiplication of vector with scalar

Let $X$ be vector space and $q : X \mapsto \mathbb {R}$ function with the following properties:

$q(x) \geq 0$ $\forall x \in X$ and $q(x)=0 \Leftrightarrow x=0$

$q(x+y) \leq q(x)+q(y)$ $\forall x,y \in X$

$q(\alpha x)\leq q(x)$ $\forall x \in X$ only if $|\alpha| \leq 1$ and

$\lim_{n\to \infty}q(x/n)=0$ $\forall x \in X$.

It is easy to notice that distance defined by $d(x,y)=q(x-y)$ induces metric topology and is translation invariant. What I had to show further that $X$ is linear topological space.

That addition is continuous, one can show easily by triangle inequality, but continuity of multiplication with scalar gave me some problems.

Define multiplication with scalar usually: $m:\mathbb{R}\times X \mapsto X$ with $m(\lambda, x) = \lambda x$. Take open neigbourhood of $\lambda x$ in $X$, say ball $B= B(\lambda x, \varepsilon)$, so we must find appropriate neighbourhoods $M,N$ of $\lambda$ in $\mathbb{R}$ and $x$ in $X$ respectively, s.t. $m(M\times N)\subseteq B$.

So, we need $q(\lambda_1 x_1 - \lambda x)<\varepsilon$ for $\lambda_1 \in M, x_1 \in N$. By some standard tricks: $q(\lambda_1 x_1 - \lambda x)=q(\lambda_1(x_1-x)+\lambda_1 x- \lambda x) \leq q(\lambda_1 (x_1 -x))+q((\lambda_1 - \lambda)x)$.

Now, it is obviously offered to choose $M=(\lambda - \frac{1}{n}, \lambda +\frac{1}{n})$, which makes the second part vanish for large $n$. But I don't know what to do with $q(\lambda_1 (x_1 -x))$.

Can someone suggest me what to do? Or am I on the right track?

The 2nd formula implies that $q(nx) \le nq(x)$ for positive integer $n$.
The 3rd formula then shows that $q(\lambda x) \le \lceil\lambda\rceil q(x)$ for any $\lambda > 0$.
• What is meaning of that symbol $\ulcorner \lambda \urcorner$ you wrote? – Gustavo Dec 11 '15 at 23:54
• $\lceil\lambda\rceil$ is the ceiling function = the least integer greater than or equal to $\lambda$. And $\lfloor\lambda\rfloor$ is the floor function = the greatest integer less than or equal to $\lambda$. And FYI, to see how some symbol is produced, right-click on the expression and select "Show Math As .. TeX commands". – Paul Sinclair Dec 12 '15 at 1:37