In the Banach space $C[0,1]$ consider the subspace
$M=\lbrace g \in C[0,1]: \int_{0}^{1}g(t)dt=0 \rbrace $
Show that $M$ is closed in $C[0,1]$ and calculate the quotient norm $(\|f+M \|)$ where $f(t)=\sin(\pi t)$ for all $t \in [0,1]$.
Probably it's easier than I think but I don't know how do this, it is obvious to see that $M$ is closed but how can I show this? And what about the norm? No idea... please help me
On
Let us denote $X=C[0,1]$. As mentioned in the comments and in the other answer, if we check that $F\colon X\to\mathbb R$ $$F\colon g\mapsto\int_0^1 g(t)dt$$ is a linear continuous functional, then $M=\ker F=F^{-1}(\{0\})$ and the continuity of $F$ implies that $M$ is closed.
We are also interested in the norm $\|f+M\|_{X/M}$ for $f(t)=\sin(\pi t)$. This norm can be expressed in several equivalent forms $$\|f+M\|_{X/M} = \inf_{g\in M} \|f-g\|_{\infty} = \inf_{h\in f+M} \|h\|_\infty.$$ You may notice that $h\in f+M$ is equivalent to $F(h)=F(f)$, this follows again from the fact that $M=\ker F$.
Specifically, for $f=\sin\pi t$ we get $F(f)=\frac2\pi$. So our problem basically reduces to finding the infimum of $\|h\|_\infty$ when we are looking at the functions $h\in C[0,1]$ such that $$\int_0^1 h(t) dt = \frac2\pi.$$
We can notice that if we choose the constant function $h_0(t)=\frac2\pi$ we have $h_0\in f+M$ and $\|h_0\|_\infty=\frac2\pi$.
We can also notice that for any function we get $$|F(h)| = \left|\int_0^1 h(t) dt\right| \le \int_0^1 |h(t)| dt \le \sup_{t\in[0,1]} |h(t)| = \|h\|_\infty.$$ This implies that $\|h\|_\infty\ge\frac2\pi$ for any $h\in f+M$.
So together we get that $$\|f+M\|_{X/M} = \frac2\pi.$$
I am just expanding @Martin Sleziak's suggestion.
We consider a functional $L:g\to \int_{0}^{1}g(t)dt$. And prove that the map $L$ is linear and continuous(which is trivial). Now the kernel of $L$ is $M$ according to definition of $M$. And The kernel of a continuous linear operator is a closed subspace(the range $\mathbb{R}$ is finite dimensional). So M is closed.
Now for the quotient norm We have to calculate $\left\lVert[\sin(\pi t)]\right\rVert_{C[0,1]/M}$ which is given by $\inf\limits_{x\in M}\left\lVert\sin(\pi t)-m\right\rVert_{\infty}$. We know $g \equiv 0 \in M$. Therefore $\left\lVert[\sin(\pi t)]\right\rVert_{C[0,1]/M} \leq \left\lVert\sin(\pi t)\right\rVert_{\infty}=1$.