How do I prove that the subset M = {f $\in$ C[0,1] : $\int_0^1f(x)dx = 0$} of C[0,1] is proximinal?

Question

I want to show that the subset $$M = \{f \in C[0,1] : \int_0^1f(x)dx = 0\}$$ is proximinal in the Banach space C[0,1](equipped with the sup norm), that is, for every g $\in$ C[0,1] there exists f$\in$M such that $\parallel$f-g$\parallel_\infty$ = dist(g,M).

Is it in this case enough to show that M is closed? Or do I need do consider something else?

Thanks for your help :)

Answer 1

Let $f=g-\int\limits_0^1g(t)\,dt.$ Then $f\in M$ and $$\| g-f\|_\infty =\left |\int\limits_0^1g(t)\,dt\right | $$ Let $h\in M.$ Then $$\|g-h\|_\infty \ge \left |\int\limits_0^1[g(t)-h(t)]\,dt\right |=\left |\int\limits_0^1g(t)\,dt\right |$$ Hence $$d(g,M)\ge \left |\int\limits_0^1g(t)\,dt\right |$$

Answer 2

In general, we can prove that if $\varphi$ -- linear bounded functional on $X$ and $M=\ker \varphi$, then for $g\in X\setminus M$ $\text{dist}(g,M)$ attains on some element of $M$ iff $\varphi$ attains it's norm.

Let $g\in X\setminus M$. It's well known, that $\forall f\in X$ have unique representation $f=h+\alpha g$, where $h\in M$ и $\alpha\in\mathbb{C}$ are different for each $f$. Then $$\|\varphi\|=\sup\limits_{f\ne0}\dfrac{|\varphi(f)|}{\|f\|}=\sup\limits_{h\in M,\;\alpha\ne0}\dfrac{|\varphi(h+\alpha g)|}{\|h+\alpha g\|}=\sup\limits_{h\in M,\;\alpha\ne0}\dfrac{|\alpha||\varphi(g)|}{|\alpha|\left\|\frac{h}{\alpha}+g\right\|}=$$$$=\sup\limits_{u\in M,\;u=-\frac{h}{\alpha}}\dfrac{|\varphi(g)|}{\left\|g-u\right\|}=\dfrac{|\varphi(g)|}{\inf\limits_{u\in M}\|u-g\|}=\dfrac{|\varphi(g)|}{\text{dist}(g,M)}.$$

In our case $\varphi(f)=\displaystyle\int\limits_0^1f(x)dx$ on $C[0,1]$ with obviously attainable norm $1$.