1
$\begingroup$

Let $A=C[0,1]$ and $B=\{f\in A|f(0)=0\}$. What is the closure of B w.r.t the $|| . ||_\infty$ norm.

I really have a very poor intuitive grasp of this problem and I'm not sure where to begin?

I know $|| f ||_\infty=\sup_{x\in[0,1]} f(x)$ and that the closure is the limit points of B

Thansk for any help

2 Answers 2

1

Alternatively:

Convergence in $\|\cdot\|_\infty$ is uniform convergence of functions, which implies pointwise convergence. Thus if there is a sequence in $B$ that converges to something outside $B$, then it would be a sequence of functions $f_n$ all with $f_n(0)=0$ which converged pointwise to an $f$ with $f(0)\ne 0$. This is clearly impossible, so $B$ is closed.

2

The map $L\colon A\to \Bbb R$, $f\mapsto f(0)$ is linear, and continuous as $|L(f)|=|f(0)|\leq \lVert f\rVert_{\infty}$. Since $B=L^{-1}(\{0\})$, $B$ is closed hence equal to its closure.