# Bounded linear operator on Banach space $C[0,1]$.

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 .

• Technically you only showed it is bounded and its norm is no more than 1, not that it is 1
– Alan
Commented Jun 10, 2021 at 7:05
• @Alan thank you ....please can you write about boundedness of inverse of T ? Commented Jun 10, 2021 at 7:13
• Sorry, been too long since my functional analysis days. I've actually just started to review my textbook for fun
– Alan
Commented Jun 10, 2021 at 7:13
• T is the composition of the Volterra operator and a multiplication operator. Volterra operator is not surjective! Its image is $C^1[0,1]$. And $C^1[0,1]$ is invariant under the multiplication operator. So T is not surjective, hence it is not invertible. Commented Jun 10, 2021 at 7:15
• @TimurBakiev sir about invertible on its range ?? I am editing it .. Commented Jun 10, 2021 at 7:19

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.
• sir I think $f(x)=x^n$ also working in place of $f_n$ given by you ? Commented Jun 10, 2021 at 13:13