# Linear transformation from the set of continuous functions $C([0,1])$ to $\mathbb{R^2}$.

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...

• Why do you say it a not an explicit function? Jan 15, 2021 at 9:57
• 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. Jan 28, 2021 at 20:40

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. $$\epsilon(f+g)=((f+g)(0),(f+g)(1))=?$$ $$\epsilon(cf)=((cf)(0),(cf)(1))=?$$, $$c\in\mathbb{R}$$