This question have been asked before but there was no explanation..
Let $X=\{f\in C[0,1] : f(0)=0\}$ with sup norm and $$Y=\{f\in X : \int_0^1f=0\}$$
Question is to show that for no $f\in X$ with $||f||=1$ we have $\rm{dist}(f,Y)<1$.
We have following result:
Let $X$ be a normed space and $T$ be a nonzero continuous functional on $X$. Then for $a\in X$ and $k\in \mathbb{C}$, show that $$\rm{dist}(a,\{x:T(x)=k\})=\frac{|T(a)-k|}{||T||}.$$
Consider the function $T:X\rightarrow \mathbb{C}$ with $$g\mapsto \int_0^1 g dx.$$ This is a continuous function. So, the result says for $k=0$ that
$$\rm{dist}(f,\{g:T(g)=0\})=\frac{|T(f)|}{||T||}.$$
As $Y=\rm{Ker}(T)$ we have $$\rm{dist}(f,Y)=\frac{|T(f)|}{||T||}.$$
So, I am left to prove that $\frac{|T(f)|}{||T||}<1$ for any $f$ with $||f||=1$
I am stuck here. I see that $|T(f)|\leq 1$. I could not find $||T||$. It is clearly $\leq 1$. But I could not say more than this.
a) The norm of $T$. Clearly, we have that $\|T\|\leq 1$. Now let $f_n$ defined by $f_n(t)=nt$ for $0\leq t\leq 1/n$, $f_n(t)=1$ if $t\in [1/n,1]$; we have $f_n\in X$, $\|f_n\|=1$. We compute easily that $\displaystyle T(f_n)=1-\frac{1}{2n}$. Hence we get that $\displaystyle 1-\frac{1}{2n}\leq \|T\|$ for all $n$, this show that $\|T\|=1$.
b) Suppose that there exists $f\in X$ with $\|f\|=1$ such that $|T(f)|=1$. Replacing eventually $f$ by $-f$, we may suppose that $\displaystyle 1=\int_0^1 f(t)dt$, ie $\displaystyle \int_0^1 (1-f(t))dt=0$. As $1-f$ is continuous and $\geq 0$, this show that $1-f=0$, in particular $f(0)=1$, contradiction as $f\in X$.