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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.

Thanks in advance.

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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.

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  • $\begingroup$ 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. $\endgroup$ – Emilio Novati Oct 22 '16 at 16:32
  • $\begingroup$ @Emilio Regardless of the intended space though the map will be of the same form. Rereading I agree with you that the intended space is not clear, but I think this still essentially answers the question. $\endgroup$ – Matt Samuel Oct 22 '16 at 16:34
  • $\begingroup$ @Emilio The edit I just made should be more accurate. $\endgroup$ – Matt Samuel Oct 22 '16 at 16:36

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