Exercise :
Let $X=C[a,b]$ be equipped with the norm $\| \cdot \|_\infty$ and $g \in X$. We define the operator $T:X \to X$ to be : $$Tf(x) = g(x)f(x), \; \forall x\in [a,b]$$ Show that $T$ is a Bounded Linear Operator $(T \in B(X))$.
Attempt :
Let $y(x),z(x) \in C[a,b]$. Then, it is :
$$T(y(x)+z(x)) = g(x)(y(x)+z(x)) = g(x)y(x)+g(x)z(x)$$ $$\implies$$ $$T(y(x)+z(x)) = Ty(x) + Tz(x), \; \forall y(x),z(x) \in C[a,b] $$
Let $\lambda \in \mathbb R$ and $f(x) \in C[a,b]$. Then :
$$T(\lambda f(x)) = g(x)[\lambda f(x)] = \lambda g(x)f(x)$$ $$\implies$$ $$T(\lambda f(x)) = \lambda f(x), \; \forall \lambda \in \mathbb R, \; \forall f(x) \in C[a,b]$$
Thus, $T$ is indeed linear.
Also, if $f(x),g(x) \in C[a,b]$ then it also is $g(x)f(x) \in C[a,b]$ and thus $T:X \to X$.
For the boundedness, I need to prove that : $$\exists M>0 : \|Tf(x)\| \leq M \|f(x)\|$$
It is :
$$Tf(x) = g(x)f(x) \Rightarrow \|Tf(x)\|_\infty = \|g(x)f(x)\|_\infty \leq \|g(x)\|_\infty \|f(x)\|_\infty $$
$$\Rightarrow$$
$$\|Tf(x)\|_\infty \leq \sup\{|g(x)| : x \in X\} \|f(x)\|_\infty $$
$$\|Tf(x)\|_\infty \leq G \|f(x)\|_\infty$$
since $g \in C[a,b]$ and thus $\exists G>0 : |g(x)| \leq G \; \forall x \in C[a,b] \implies G = \sup\{|g(x)| : x \in X\}$.
Question : Is my proof correct and rigorous enough ?