Unit Ball in 1 norm is open in ($C[0,1] , || \quad ||_{\infty}$)

Question

Claim $B_1(0,1) := \{ f \in C[0,1] ; ||f||_{1} < 1 \} $ in $(C[0,1],||\quad || _1)$ is open in $(C[0,1],||\quad || _{\infty}).$

We need to take any $f \in B_1(0,1),$ and we have to find an $\epsilon$ such that $B_{\infty}(f,\epsilon) \subset B_1(0,1)$

I don't have any idea of finding that $\epsilon $

Answer 1

If $g\in B_\infty(f,\epsilon)$ then $||f-g||_\infty<\epsilon$ and consequently, $||f-g||_1=\int_0^1|f-g|dx<\epsilon$.So $||f||_1-\epsilon \leq||g||_1\leq||f||_1+\epsilon$.

Therefore if you take $\epsilon=\frac{1-||f||_1}{2}$, you obtain $B_\infty(f,\epsilon)\subset B_1(0,1)$.