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?
Best Answer
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 $\|T\|=2$.