Find the norm of the following operator. 
Take $X=C([0,1])$ with the uniform norm, $\|f\|=\sup_{x\in[0,1]}|f(x)|$, and define the operator $T:X\to X$ by,
  $$T(f)(x)=f(x)-\int_0^1f(s)ds$$
  Find $\|T\|$.

I was hoping to solve this problem using directly the definition of the operator norm, namely,
$$\|T\|=\sup\{\|Tx\|:x\in X,\|x\|\le1\}=\{\frac{\|Tx\|}{\|x\|}:x\in X, x\neq0\}$$
However I am not sure how to begin. I have already shown that $T$ in this case is a bounded linear operator, being bounded above by $k=2$, in the sense that,
$$\|Tf\|\le2\|f\|,\,\,\,\forall f\in X$$
The solution we done in class involved constructing a sequence of functions $(f_n)_n^\infty\subset X$ such that $Tf_n\to2$. Is there any way to find $\|T\|$ without constructing such a sequence, by using the definition of the operator norm?
 A: Consider the functions $f_n$ given by $f_n(x)=-1$ if $x\leq 1-1/n$ and $f_n(x)=2nx-2n+1$ if $1-1/n\leq x\leq 1$ (i.e. the linear interpolation between $(1-1/n,-1)$ and $(1,1)$). Then we have
$$Tf_n(x)=f_n(x)-\int_0^1 f_n(s) ds=f_n(x)+1-1/n$$
and $\|Tf_n\|=2-1/n$. So indeed $\|T\|=2$.
A: Let $f$ be any continuous real function such that
$$
               \inf_{x\in[0,1]}f(x)= -1,\;\;\;\sup_{x\in [0,1]}=1
$$
Then $\|f\|=1$. If the integral average $\int_{0}^{1}f(x)dx$ of $f$ is anything other than $0$, then $\|f\| < \|f-\int_{0}^{1}f(t)dt\|$. You can imagine that it is possible to choose $f$ so that the integral average is arbitrarily close to $1$, even thought $f$ assumes values of $\pm 1$; for such a function $Tf=f-\int_{0}^{1}f(t)dt$ will have a norm that can be made arbitrarily close to $2$.
A: Consider the function $$\begin{array}{l|rcll}
f_n : & [0,1] & \longrightarrow & \mathbb R \\
    & x & \longmapsto & 1 - 2nx & \text{ for } 0 \le x \le \frac{1}{n}\\
    & x & \longmapsto & -1 & \text{ for } \frac{1}{n} < x \le 1
\end{array}$$
You have $\Vert f_n \Vert = 1$ for all $n \in \mathbb N$ and $$\int_0^1 f_n(t) dt=-(1-\frac{1}{n})$$ Hence $$T (f_n)(x)=f_n(x)+\left(1-\frac{1}{n}\right)$$ In particular $T(f_n)(0)=2-\frac{1}{n}$ tends to $2$ as $n \to +\infty$. Which implies that $\lim\limits_{n \to +\infty} \Vert T(f_n) \Vert = 2$ and $\Vert T \Vert = 2$
