Show $F_1$ is a continuous linear functional Let $(X,||\cdot ||)=(C[0,1],||\cdot ||_{\infty})$ and $F_1 :X \rightarrow \mathbb R$ be defined by $$F_1 (f)=\int _{1/2}^{3/4} f(t) dt$$ Show $F_1$ is a continuous linear functional.

So we need to show it is a linear functional first and then show it is continuous. By our lecture notes, if we show it is continuous at $0$, then it is continuous everywhere. So then the question is done.
Linear functional: Let $a \in \mathbb R$ and $f,g \in X$. Then $$F_1(af+g)=\int _{1/2}^{3/4} (af+g)(t) \, \, dt = \int _{1/2}^{3/4} af(t)+g(t) \, \, dt=\int _{1/2}^{3/4} af(t) dt + \int _{1/2}^{3/4} g(t) dt$$ $$ = a \int _{1/2}^{3/4} f(t) dt + \int _{1/2}^{3/4} g(t) dt = aF_1 (f) + F_1 (g)$$ So it is a linear functional.
Prove that it is continuous at $f=0$: Need to prove $$\forall \varepsilon >0, \, \exists \delta >0: \, ||F_1(f)-F_1(0) ||_{X^{*}} < \varepsilon \, , \text{whenever} \,  ||f-0||_{\infty}$$ So $$||F_1 (f) ||_{X^{*}} = \sup _{f \in X, \, \, ||f||_{\infty}<\delta} \bigg( | \int _{1/2}^{3/4} f(t) dt | : ||t ||_{\infty}\leq 1 \bigg) = \sup _{f \in X, \, \, ||f||_{\infty}<\delta} \bigg(  \int _{1/2}^{3/4} |f(t)| dt  : ||t ||_{\infty}\leq 1 \bigg) \leq  \int _{1/2}^{3/4} \delta dt  =\frac14 \delta < \delta = \varepsilon$$ We let $\delta = \varepsilon$.
Is this correct? I have a feeling I made some notation errors somewhere.
 A: For any $\;\epsilon>0\;$ choose $\;\delta=4\epsilon\;$ , so that
$$||f-0||_\infty\iff ||f||_\infty<\delta\implies\left\|\int_{1/2}^{3/4}f(t)dt\right\|\le\int_{1/2}^{3/4}||f(t)||dt\le\delta\int_{1/2}^{3/4}dt=\frac14\delta=\epsilon$$
A: To prove the norm of $F_1$ is $1/4$, we can go back to the definition $$\| F_1 \|_{X^*} = \sup \{ \lvert F_1(f) \rvert : \| f \|_\infty \le 1\}.$$ For $f \in X$, with $\| f \|_\infty \le 1$, we have $$\lvert F_1(f) \rvert = \left \lvert \int^{3/4}_{1/2} f(x) dx \right \rvert \le \int^{3/4}_{1/2} \lvert f(x) \rvert dx \le \int^{3/4}_{1/2} 1 dx = \frac 1 4.$$ Since this holds for all $f$ with $\| f \|_{\infty} \le 1$, we have $\| F_1 \|_{X^*} \le \frac 1 4.$ 
Now define $$g(x) = \left \{\begin{matrix}2x, & 0 \le x \le 1/2, \\ 1, & 1/2 \le x \le 3/4, \\ 4(1-x), & 3/4 \le x \le 1. \end{matrix} \right.$$ Then $g$ is continuous so $g \in X$; also $\| g \|_\infty = 1$. Since $\| F_1 \|_{X^*}$ is the supremum of $\lvert F_1(f) \rvert$ over such functions, in particular. we must have $\| F_1 \|_{X^*} \ge \lvert F_1(g) \rvert$. We see $$\lvert F_1(g) \rvert = \left \lvert \int^{3/4}_{1/2} g(x) dx \right \rvert$$ but between $1/2$ and $3/4$, this $g$ is just the constant $1$. Thus $$\lvert F_1(g) \rvert = \left \lvert \int^{3/4}_{1/2} 1 dx \right \rvert = \frac 1 4.$$ Hence, $\| F_1 \|_{X^*} \ge \frac 1 4.$ Thus $\| F_1 \|_{X^*} = \frac 1 4.$
