Question : Let $C[0,1]$ be the Banach space of continuous functions on $[0,1]$ with supremum norm . Discussed about boundedness of the operator $T$ on $C[0,1]$ and it’s inverse, where $T$ is given by $T(f)(x)=\int_0^x xf(t)dt$.
I tried as below
$$\|T(f)\|=sup |\int_0^x xf(t)dt|\leq sup\int_0^x |xf(t)|dt\leq \|f\|sup|x|=\|f\|$$
So $\|T(f)\|\leq \|f\|$, which implies that $T$ is bounded with norm $1$. One more thing I noted that $T$ is one one so invertible, but I am unable to tell about boundedness of $T^{-1}$ on range $R$ of $T$. Please help me to tell about boundedness of $T^{-1}$ on $R$, and tell me the way I checked bounded of $T$ is correct? Thank you .
Best Answer
You have proved that $T$ is bounded with $\|T\| \leq 1$.
Let $f_n(x)=nx$ for $0\leq x \leq \frac 1 n$ and $f_n(x)=n(\frac 2 n -x)$ for $\frac 1 n \leq x \leq \frac 2 n$. Then $f_n$ is continuous and $\|f_n\|=1$ for each $n$. You can easily check that $\|Tf_n|| \leq \frac 2 n$. So there cannot be any positive constant $c$ such that $\|Tf \| \geq c \|f\|$ for all $f$. It follows that $T$ does not have a continuous inverse.