Scalar multiplication and continuity on a metric space

Question: Using the metric space definition of continuity, prove that scalar multiplication by a fixed $\alpha \in \mathbb{R}$ is continuous at $\vec{0}$.

The question makes no reference as to what f is but it is likely that f is a function in the continuous space map $C\left ( \left [ a,b \right ],\mathbb{R} \right )$.

Definition:

Let $\left ( X,d \right ) and \left ( Y,e \right )$ be metric spaces. Let $f:\left ( X,d \right )\rightarrow \left ( Y,e \right )$ be a function and let $a \in X$. f is continuous at a if for every $\epsilon >0$, there exists $\delta >0$ s.t $f\left ( B_{\delta } \left ( a \right )\right )\subseteq B_{\epsilon }\left ( f\left ( a \right ) \right )$

Any hint is appreciated. It is quite unclear to me what the question meant by "prove that scalar multiplication by a fixed $\alpha \in \mathbb{R}$ is continuous at $\vec{0}$."

Any clarification is also appreciated.

The function $f:X\to X$ is given by $$f(x)=\alpha x$$ for some fixed $\alpha$, whatever your space $X$ is. This is the meaning of the scalar multiplication function.
• If I well understand the OP asks for a function $F:C[a,b]\to C[a,b]$ such that $F(f)=\alpha f$. But is is not clear what is the metric ( or norm) on the space of functions. – Emilio Novati Oct 22 '16 at 16:32