find norm of linear operator

Question

let $\mathbf{C}[0,1]$ be the space of continuous real-valued functions on $[0,1]$ with sup norm and $\mathbf{T}: \mathbf{C}[0,1] \rightarrow \mathbb{R}$ be linear operator such that :

$\mathbf{T}(\mathbf{f})=\int_{0}^{1} \mathbf{f}(\sqrt{\mathbf{x}}) \mathrm{d} \mathbf{x} \quad,(\mathbf{f} \in \mathbf{C} [0,1])$

now find $\|\mathrm{T}\|$ ?

\begin{array}{l} 1){\|\mathrm{T}\|=\infty} \\ 2){\|\mathrm{T}\|=\mathrm{2}} \\ 3){\|\mathrm{T}\|=\frac{1}{\mathrm{2}}} \\ 4){\|\mathrm{T}\|=1} \end{array}

we know $$\|T\|=\sup\{\|Tx\|:x\in X,\|x\|\le1\}=\{\frac{\|Tx\|}{\|x\|}:x\in X, x\neq0\}$$ and $\|f(x)\| =\sup_{x\in [0,1]}|f(x)|=\max_{x\in [0,1]}|f(x)|$ for $f(x)=1 $ we have $\|T(f(x))\| =1 $ so option $3$ is false

Answer 1

What you did show that the norm of $T$ is bigger or equal to one. Using $\lvert f\left(\sqrt x\right)\rvert\leqslant \lVert f\rVert_\infty$ for all $x$, we can deduce that the norm of $T$ is smaller or equal to one.

Answer 2

You have $|Tf| \le \int_0^1 \|f\| dx = \|f\|$, so $\|T\| \le 1$. Let $f(x) = 1$ for all $x$ to get $|Tf| = 1$ hence $\|T\| =1 $.

Answer 3

Call $t=\sqrt{x}$. Then your Intégral becomes $$ I=2\int_0^1 tf(t)dt $$ For $f$ with $|f|\le 1$ you have $$ |I|\le 2\int_0^1 tdt\le 1 $$ Hence the norm is 1.