Proving that $\gamma:[0,1]\to V$ is continuous

Question

I'm trying to prove that if $(V,|\cdot|)$ is a normed vector space over $\mathbb C$ and $x,y\in V$, then the map $$\gamma:[0,1]\to V$$ defined by $\gamma(t)=x(1-t)+yt$ is continuous, where $[0,1]$ is considered with the subspace topology induced by the Euclidean topology in $\mathbb R$, and $V$ with the topology induced by the metric $(a,b)\mapsto|b-a|$

I was trying to prove that the pre-image of $B_r(p)$ for any $r>0$ and $p\in V$ is an interval in $[0,1]$ and I was wondering if there is any easier way of doing this apart from the definition of continuity. How would you prove this?

Answer 1

We have $d(t,s) := \gamma(t) - \gamma(s) = -x(t-s) + y (t-s)$, thus $$\lvert d(t,s) \rvert \le \lvert t-s \rvert \cdot \lvert-x \rvert + \lvert t- s \rvert \cdot \lvert y \rvert = \lvert t-s \rvert (\lvert x \rvert + \lvert y \rvert)$$ which proves that $\gamma$ is (uniformly) continuous.