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.

For any $\;\epsilon>0\;$ choose $\;\delta=4\epsilon\;$ , so that
$$||f-0||_\infty\iff ||f||_\infty<\delta\implies\left\|\int_{1/2}^{3/4}f(t)dt\right\|\le\int_{1/2}^{3/4}||f(t)||dt\le\delta\int_{1/2}^{3/4}dt=\frac14\delta=\epsilon$$
