Consider the operator $T : C[0,1] \to C[0,1] $ ($C[0,1]$ has the maximum metric) defined as $$T(f)=\int_0^x (x-t)f(t)dt$$ Prove that $T$ is well defined.
So essentially what I want to prove is that $Tf \in C[0,1]$.
To do that I think I need to prove that $Tf$ is continuous and bounded.
$Tf$ is continuous since it's an integral of a continuous function.
To prove that it's bound I work like this $$\| Tf \| =\| \int_0^x (x-t)f(t) dt \| \leq \| f (x) \| \| \int_0^x (x-t)dt \| \leq K \max_{x \in [0,1] } |\frac{x^2}{2} | \leq \frac{K}{2} $$ Since $f$ is bounded (i.e $\| f \| \leq K$) and $\frac{x^2}{2} $ has a maximum of $\frac{1}{2} $ on $[0,1]$.
So $Tf \in C[0,1] $.
Is my proof correct?
You have: $$\forall x \in [0,1], T(f)(x)=\int_0^x (x-t)f(t)dt=x\int_0^x f(t)dt-\int_0^x tf(t)dt$$
By hypothesis, $f(t)$ and $tf(t)$ are continuous on $[0,1]$, therefore $T(f)(x)$ is continuous on $[0,1]$. So $T(f)$ is well-defined.
Not sure why you would want to show that $f$ is bounded, though. It is true because we now know that $T(f)$ is continuous on $[0,1]$ (extreme value theorem) but it's not needed to show that $T(f)$ is defined.