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Let $\varepsilon : C([0,1]) \to \mathbb{R^2}$ be a mapping $$f\mapsto (f(0), f(1))$$ show that $\varepsilon$ is a linear transformation from the set of continuous functions $C([0,1])$ to $\mathbb{R^2}$.

For $\varepsilon$ to be linear transformation it would need to satisfy $\varepsilon(x+y) = \varepsilon(x) + \varepsilon(y)$ and $\varepsilon(cx) = c\varepsilon(x)$. However, I'm not sure how to approach this since we're not given any excplicit function to show this? Any hints would be appreciated...

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  • $\begingroup$ Why do you say it a not an explicit function? $\endgroup$
    – Kavi Rama Murthy
    Jan 15, 2021 at 9:57
  • $\begingroup$ Here, rather than $x,y$, the vectors in $C([0,1])$ would perhaps be better represented as variables $f,g$. Remember what the domain vector space is and how vector addition and scalar multiplication are defined there. $\endgroup$
    – hardmath
    Jan 28, 2021 at 20:40

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You need just the definition of function addition and multiplacation of function by a scalar. In $C([0,1])$ functions have both same domain and codomain, so you have permission for addition and scalar multiplication. $\epsilon(f+g)=((f+g)(0),(f+g)(1))=?$ $\epsilon(cf)=((cf)(0),(cf)(1))=?$, $c\in\mathbb{R}$

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