Open unit ball in integral norm is open in supremum norm

Question

I have to show the following

I know I have to find an $\epsilon$ such that $B_\infty(f,\epsilon)$ is in $B_1(0,1) \forall f$ . I just cannot figure out what this epsilon could be. Also I don't know how to use that hint given in the question. I think I am missing something.

Kindly help thanks and regards

Answer 1

Let $f \in B_1(0,1)$. Set $\delta=\min\{ ||f||_1 , 1 - ||f||_1 \}$.

Now conside $B_0(f,\delta/2) = \{ g\in C[0,1]: ||f-g||_\infty < \delta /2 \}$.

Now $g\in B_0(f, \delta/2)$, then $||f-g||_1<||f-g||_\infty<\delta/2$

$||g||_1 \le ||f||_1+||f-g||_1 < ||f||_1 + \delta/2 < 1$

Hence, $g\in B_1(0,1) \Rightarrow B_0(f,\delta/2) \subset B_1(0,1)$

Therefore, $B_1(0,1)$ is open in $(C[0,1],||.||_\infty)$.