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 $||x-y|| \geq 1 ~~\forall y \in Y$
 A: I am just studying Functional Analysis, so I wish to have my solution checked. 
Since $X$,$Y$ are subspaces of $[0,1]$ so is $X \cap Y$. Now $X \cap Y \subseteq X$ is closed subspace in $X$.  Since if we choose a Cauchy Sequence $\{p_n\}$ in $X \cap Y$, $\{p_n(t)\}$ is bounded for every $t \in [0,1]$ and since $\{p_n\}$ is Cauchy, $\{p_n(t)\}$ converges for every $t\in [0,1]$. 
$$p(t)=\lim\limits_{n \to \infty}p_n(t)$$
Since $p_n(0)=0$ for every $n$, we have $p(0)=0$ and since $p_n \in X \cap Y \subseteq C[0,1]$, $p_n$ is bounded and we have $\lim \limits_{n \to \infty} \int _0^1 p_n(t)dt=\int_0^1\lim p_n(t)dt=0$ and hence $X \cap Y$ is closed. 
Then by Riesz's lemma,  for $0 < \epsilon < 1$, we are guaranteed of an $x \in X$ such that 
$$||x-y||\ge 1-\epsilon$$ such that for any $y$ satisfying $y(0)=0, \int_0^1y(t)dt=0$.
Edited the answer before the question was altered to show a weaker existence of the function satisfying the conditions. For a stronger condition, discussions in comments shows that the underlying space should be reflexive for a stronger form of Riesz's Lemma to hold. 
A: 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:
$$|m| \geq 1$$.
