Trouble with simple notation in proving linearity of an operator. Consider the operator $T\colon \mathcal C[0,1]\to \mathcal C[0,1]$ defined by
$$ Tf(x) = \int_0^x f(t) \, dt, \quad \forall f \in \mathcal C[0,1].$$
To prove its linearity, consider two scalars $\alpha,\beta \in \mathbb K$ and two functions $f,g \in \mathcal C[0,1]$, as arbitrary as possible. Then,
\begin{equation*}
    T(\alpha f+\beta g)\color{red}{(x)} = \int_0^x (\alpha f + \beta g)(t)\, dt = \int_0^x \alpha f(t) + \beta g(t) \, dt = \alpha \int_0^x f(t) \, dt + \beta \int_0^x g(t) \, dt = \alpha Tf\color{red}{(x)} + \beta Tg\color{red}{(x)}.
\end{equation*}
My question. Does my notation in $\color{red}{red}$ makes sense? Somehow, in my head it would make more sense if I just wrote $T(\alpha f + \beta g)$ but, at the same time, according to the definition it doesn't make that much sense. I get even more confused when it comes to the final conclusion: We just proved that
$$ T(\alpha f + \beta g) = \alpha Tf + \beta Tg, \quad \forall \alpha,\beta \in \mathbb K, \, \forall f,g \in \mathcal C[0,1].$$
Here, I didn't use the notation in $\color{red}{red}$ since it makes zero sense to me.
Thanks for any help in advance.
 A: It would be better to put brackets around $Tf$. So, it should look like $$(Tf)(x)=\int_{0}^x f(t)dt$$ This is because the function $T$ takes a function $f\in C([0, 1])$ to the function $Tf : [0, 1]\to K$ defined by $$(Tf)(x)=\int_{0}^x f(t)dt$$
When proving the linearity of $T$, the usage of $x$ is vague. The $x$ should be placed like $$(T(\alpha f+\beta g))(x)$$ Because the $x$ corresponds to the function $T(\alpha f + \beta g)$ and not $(\alpha f+\beta g)$. After the usage of $x$ is cleared up, you should include the $x$ as we have to prove that the function $T(\alpha f + \beta g)$ is equal to the function $\alpha(Tf)+\beta(Tg)$. And when proving that two functions $h$ and $j$ are equal, one often proves that $h(x)=j(x)$ for all $x$. Then using the linearity of the integral we conclude that $$(T(\alpha f+\beta g))(x)=\alpha(T f)(x)+\beta(T g)(x)\text{ for all }x\in [0, 1]$$
Also, it is vague to write $\alpha Tf(x)$ as the usage of $x$ is vague. So it is better to write $\alpha(Tf)(x)$. The same goes for $\beta Tg(x)$. It should be written as $\beta (Tg)(x)$.
The above expression is a bit convoluted. Breaking up the above expression in terms of additivity and homogeneity, we have shown that $$(T(f+g))(x)=(Tf)(x)+(Tg)(x)\text{ for all }x\in [0, 1]$$ and $$(T(\alpha f))(x)=\alpha (Tf)(x)\text{ for all }x\in [0, 1] \text{ and for all }\alpha \in K$$ It doesn't matter whether we include the $\alpha $ inside the brackets $(Tf)$ as the usual definition of scalar multiplication of a function is defined by $$(cf)(x)=cf(x)$$
Hope this clears things up!
