Define a metric space $C(K)=\left \{ f: K\rightarrow \mathbb{R} \right\} $ , where $f$ is continuous function on K.
Define $\delta :C([0,1],\mathbb{R})\rightarrow \mathbb{R}$ by $\delta (f)=f(0)$.
Prove that $\delta $ is linear and continuous.
(My attempt) I think there's a little wrong expression in this problem because of the expression $f(0)$. I think the expression should be $f((0,0))$,so I used this expression.
First I showed the linearity. $\delta (f+g)=(f+g)((0,0))=f((0,0))+g((0,0))=\delta (f)+\delta (g)$. For $c\in \mathbb{R}$,$\delta (cf)=cf((0,0))=c\delta (f)$.
Next I wanted to show the continuity of $\delta $, but it seems like $\delta $ is continuous by the definition.I learned Arzela-Ascoli theorem in class recently, and learned the new notation $C(K)=\left \{ f: K\rightarrow \mathbb{R} \right\} $ , where $f$ is continuous function on K. I think I'm confused with the notation.
Hint. The function $\delta$ is from the metric space $C([0,1])$ endowed with the uniform distance $d(f,g)=\max_{x\in [0,1]}|f(x)-g(x)|$ to the metric space $\mathbb{R}$ with the euclidean distance $d'(x,y)=|x-y|$. Now, note that $$d'(\delta (f),\delta (g))=|\delta (f)-\delta (g)|=|f(0)-g(0)|\leq d(f,g).$$ P.S. The notation $\delta(f)=f(0)$ is fine: given a function $f:[0,1]\to\mathbb{R}$, $\delta(f)$ is the value of $f$ at the point $0$.