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?
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 $$ \inf\{||g-f||_{\infty} | g\in Y \} = \left|\int_0^1 f(x)dx \right|<1. $$