Show $T$ is a bounded linear operator

Question

Let $T:C[0,1]\rightarrow C[0,1]$ be defined as: $$(Tf)(x)=\int_{0}^{1}xyf(y)\,\mathrm{d}y.$$

Show $T$ is a bounded linear operator and calculate its norm.

My idea: First I know that $\Vert f\Vert=\sup_{y\in[0,1]} |f(y)|$ and $f$ is defined and continuous on $[0,1]$. I feel that the integral is clearly not ill-defined since: $$\int_{0}^{1}xyf(y)\,\mathrm{d}y\leq x\int_{0}^{1}y\cdot\sup_{z\in[0,1]}|f(z)|\,\mathrm{d}y=x\sup_{z\in[0,1]}|f(z)|$$ (not sure about that) Now, if $\Vert f\Vert\neq 0$, we can find the norm of the operator by applying the following formula:
$\sup\frac{\Vert Tf\Vert}{\Vert f\Vert}=\sup\frac{\sup\left|\int_{0}^{1}xyf(y)dy\right|}{\sup|f|}$, but where does it lead?

Answer 1

Well, as you (nearly) already noticed $$ \left| \int_0^1 xy f(y) \,\mathrm{d}y \right| \le x \|f\|_\infty \int_0^1 y \,\mathrm{d}y = \frac{x}{2} \|f\|_\infty, $$ hence $$\|Tf\|_\infty \le \frac12 \|f\|_\infty.$$ Therefore $T$ is bounded with $\|T\| \le \frac12$.

For $f \equiv 1$ you get $(Tf)(x) = \frac{x}{2}$, hence $\|Tf\|_\infty = \frac12 \|f\|_\infty$, so $\|T\| \ge \frac12$.

Answer 2

If $x\in[0,1]$ and $\lVert f\rVert=1$, then\begin{align}\bigl\lvert(Tf)(x)\bigr\rvert&=\left\lvert\int_0^1xyf(y)\,\mathrm dy\right\rvert\\&\leqslant x\int_0^1y\bigl\lvert f(y)\bigr\rvert\,\mathrm dy\\&\leqslant x\int_0^1y\,\mathrm dy\\&=\frac x2.\end{align}Therefore, $\lVert Tf\rVert\leqslant\frac12$. Actuallly, $\lVert Tf\rVert=\frac12$; just see what happens if $f=1$.