Let $C[0,1]$ be the Banach space of real valued continuous functions on $[0,1]$ equipped with the supremum norm. Define $T: C[0,1] \to C[0,1] $ by $$T(f(x))=\int_{0}^{x}xf(t)dt.$$
Let $R(T)$ denote the range of $T.$
Consider the following statements.
P: $T$ is a bounded linear operator.
Q: $T^{-1}:R(T) \to C[0,1]$ exists and is bounded.
My attempt: For any $f \in C[0,1],$ we have that
$$ || T(f(x))||=\sup_{x \in [0,1]} \bigg| \int_{0}^{x}xf(t) dt\bigg| \leq \sup_{x \in [0,1]} |f(t)| = ||f||.$$
The inequality in the above equation is because we are dealing with $x \in [0,1].$
So, $T$ is a bounded linear operator. So, statement P is TRUE.
As for Q, I shall quote a result.
Let $T:X \to Y$ be a linear operator between two normed spaces $X,Y,$ and let $T$ be onto. Then, $T^{-1}$ exits and is continuous if and only if there exists $c>0$ such that $c||x|| \leq ||T(x)||$ for each $x \in X.$
One can easily see that $||T|| \leq 1,$ with equality achieved for $f(x)=1.$
Now, set $X=C[0,1],Y=R(T).$ For, $c=1,$ we have satisfied all properties of the above quoted result. So, $T^{-1}$ exits and it is continuous and consequently bounded.
So, statement Q is also TRUE.
But the answer key says that P is TRUE and Q is FALSE.
I am confused as I think I have applied things correctly.
Please help me.
Best Answer
Let $f_n(x)=1$ for $0 \leq x \leq \frac 1 n$, $0$ for $x >\frac 2n$ and let $f_n$ have a linear graph between $\frac 1n$ and $\frac 2n$. You can esaily check that $\|Tf_n\| \to 0$ but $\|f_n\|=1$ for every $n$. Hence, there is no positive number $c$ such that $c\|f\| \le \|Tf\|$ for all $f$.