Find the norm of $t+ M$ in the quotient space $C[0,1]/M$

Question

Let $X=C[0,1]$ and $M=\{f \in C[0,1] : f(0)=f(1)=0\}$ construct the space $X/M$ and say what is the norm of the element of $X/M$ that contains $f(t)=t$

First, let $f,g \in C[0,1]$ such that $f-g \in M$ then $(f-g)(1)=(f-g)(0)=0$ these implies that $f(1)-g(1)=0$ and $f(0)-g(0)$ therefore $f(0)=g(0)$ and $f(1)=g(1)$ so the "class" of $f$ in $X/M$ is $$[f]=f+M=\{g \in C[0,1] : g(0)=f(0),g(1)=f(1)\}$$ and these are the characterizations of elements of $X/M$
Now $||t+M||=\inf_{g \in m} ||f-g||$ i try to find a sequence $\{g_n\} \in M$ such that $\lim_{n \to \infty} g_n=f$ but i dont sure to these is the righ way, any hint or suggestion i will very grateful.

Answer 1

$\|f+g\|\geq |f(1)+g(1)|=1$ if $g \in M$. Hence, $\inf_{g \in M}\|f+g\| \geq 1$. If $g_n(t)=1-t$ for $\frac 1 n \leq t \leq 1$ and $g_n(t)=(n-1)t$ for $0 \leq t \leq \frac 1 n$ then $g_n \in M$ and $\|f+g_n\| \to 1$. Hence, $\|[f]\|=1$.

EDIT: As pointed out in the comment below, we can also take $g=0$.

Answer 2

As was observed in the previous answer $$\inf_{g\in M}\|t+g\| \ge 1$$ Next for $g=0$ we have $\|t+g\|=\|t\|=1.$ Hence $\|[t]\|=1.$

More generally $$\|[f]\|=\max\{|f(0)|,|f(1)|\}\qquad (*)$$ Indeed for $g\in M$ we have $$\|f+g\|\ge \max\{|f(0)+g(0)|,|g(1)+f(1)|\}=\max\{|f(0)|,|f(1)|\}$$ Next for given $\varepsilon >0$ there is $n$ such that $|f(u)-f(v)|<\varepsilon$ for $|u-v|\le n^{-1}.$ Let $$ g(x)=\begin{cases} -f(x)&n^{-1} \le x\le 1-n^{-1} \\ -f(n^{-1})\,nx ,& 0\le x\le n^{-1} \\ -f(1-n^{-1})\,(nx-n) & 1- n^{-1}\le x\le 1\end{cases}$$ Then $$f(x)+g(x)=\begin{cases} 0 & n^{-1} \le x\le 1-n^{-1} \\ f(x)-f(n^{-1})\,nx ,& 0\le x\le n^{-1} \\ f(x)-f(1-n^{-1})\,(nx-n) & 1- n^{-1}\le x\le 1\end{cases}$$ Thus $$\|f+g\|\le \max\{|f(0)|,|f(1)|\}+2\varepsilon$$ As $\varepsilon>0$ is arbitrary we get $$\inf_{g\in M}\|f+g\|\le \max\{|f(0)|,|f(1)|\}$$ Finally we obtain $(*)$.