Find continuos function $x$ with $x(0)=0$ such that $\|x-y\| \geq 1$ where $y(0)=0$ and $\int_0^1 y(t) dt = 0$ Consider $C[0,1]$ space of continuos functions with the uniform norm. Let $X = \{ x \in C[0,1]: x(0)=0\}$ and $Y = \{ y \in X: \int_0^1 y(t) dt = 0 \}$ subspaces of  $C[0,1]$. How can I show that $\exists x \in X$, $\|x\|=1$, such that $$\|x-y\| \geq 1 ~~\forall y \in Y?$$
Edit: Sorry guys, the question is: How can I prove that doesn't exists $x \in X$, such that $||x||=1$ and $||x-y|| \geq 1 ~~\forall y \in Y$

Let us suppose there exists such a function $x$. We define $m$ by:
$$m=\int_0^1 x(t) dt$$
the idea is to show that $|m|\geq1$ which is in contradiction with $x$ continuous, $x(0)=0$ and $\|x\|_\infty=1$.
Let $\epsilon >0$. There exist $g_\epsilon \in X$ such that:
$$\int_0^1 g_\epsilon(t) dt=1-\epsilon, \, \|g_\epsilon\|=1$$
(take for example $g_\epsilon(t)=1$ for $t>2 \epsilon$ and $g_\epsilon(t)=\frac{t}{2\epsilon}$ for $t \leq 2 \epsilon$).
Then:
$$\tilde{x}_\epsilon(t)=x(t)-\frac{m}{1-\epsilon} g_\epsilon(t)$$
is in $Y$ as it is in $X$ (as a linear combination of function in $X$) and :
$$\int_0^1\tilde{x}_\epsilon(t)dt=\int_0^1x(t)dt-\frac{m}{1-\epsilon} \int_0^1 g_\epsilon(t) dt=m-\frac{1-\epsilon}{1-\epsilon} m=0$$.
So:
$$\|x-\tilde{x}_\epsilon\|=\left\|\frac{m}{1-\epsilon} g_\epsilon \right\| \geq 1$$ i.e:
$$\frac{|m|}{1-\epsilon} \geq 1$$.
So for any $\epsilon >0$:
$$|m| \geq 1-\epsilon$$
and thus:
$$|m| \geq 1$$.
