Banach spaces exercise Let $(C[0,1],\lVert\cdot\rVert_{\infty})$ be the set of continuous functions in $[0,1]$, and consider $X=\{f\mid f(0)=0\}$ and  $Y=\{f \in X\mid \int^1_0f(x)\,\textrm{d}x=0\}$, subspaces of $C[0,1]$.

Prove that for all $f \in X$ such that $\lVert f\rVert_{\infty}=1$ we have that  $\inf\{\lVert g-f\rVert_{\infty}\mid g \in Y \}<1$.

I proved that $Y$ is a proper closed subspace of $X$, thus is Banach because I also proved that $X$ is Banach.
We have that  $\inf\{\lVert g-f\rVert_{\infty}\mid g \in Y \} \leqslant \lVert f-0\rVert _{\infty}=1$.
Now if I assume that  $\inf\{\lVert g-f\rVert_{\infty}\mid g \in Y \} =1$, can someone help me to derive a contradiction?
 A: Let $f(0)=0$ and $||f||_{\infty}=1$. Put $\beta=\int_0^1 f(x) dx$. Then $|\beta|<1$. 
It is possible to find an explicit evaluation of $\inf\{||g-f||_{\infty} | g\in Y \}$. Note that for any $g\in Y$, 
$$
\left| \int_0^1 (f(x)-g(x)) dx \right| \leq ||g-f||_{\infty}. 
$$
This gives 
$$
\left|\int_0^1 f(x)dx \right| =|\beta|\leq \inf\{||g-f||_{\infty} | g\in Y \}.
$$
To prove reverse inequality, we look for $g(x)$ in the form of 
$$g(x)=f(x)-h(x)\beta.$$
Then we require
$$
h(0)=0, \ \ \int_0^1 h(x) dx =1, \textrm{ and  } \ \sup_{x\in [0,1]}|h(x)\beta|<1.
$$
In case $\beta=0$, we put $g(x)=f(x)$. Otherwise, we look for a continuous function $h(x)$ satisfying 
$$
h(0)=0, \ \ \int_0^1 h(x)dx=1 , \textrm{ and  } \  \sup_{x\in [0,1]}|h(x)|<\frac1{|\beta|}.
$$
Let $\epsilon>0$. By using $h(x)$ defined by
$$
h(x) = x^{\epsilon} (1+\epsilon),$$
we obtain 
$$
h(0)=0, \ \ \int_0^1 h(x) dx = 1, \ \textrm{ and } \ \sup_{x\in [0,1]} |h(x)| = 1+\epsilon.
$$
Then $||g-f||_{\infty} = \sup_{x\in [0,1]} |h(x) \beta| = (1+\epsilon)|\beta|.$
Since we may take $\epsilon$ arbitrarily small, we obtain the result
$$
\inf\{||g-f||_{\infty} | g\in Y \} = \left|\int_0^1 f(x)dx \right|<1.
$$
