This question is similar to this one but the only response was pointing out mistakes in the solution.
My goal is to determine whether the operator $T: C[0,1] \to C[0,1]$ defined by $Tx = \int_{0}^{t} x(\tau)\,d\tau$ is bounded.
I know that by definition, for an operator to be bounded, the following inequality must hold for a real number $c > 0$: $||Tx|| \leq c ||x||$.
Therefore, I have: $||Tx|| = ||\int_{0}^{t} x(\tau)\,d\tau || \leq \int_{0}^{t} |x(\tau)|\,d\tau \leq max\,|x(\tau)| t = ||x|| t$. Hence $||Tx|| \leq ||x|| t$ and since $t$ is not fixed, then the operator is not bounded.
My question is on the validity of bringing down the integration variable. Had the integral been $\int_{0}^{1} |x(\tau)|\,d\tau$, I would have felt comfortable stating $\int_{0}^{1} |x(\tau)|\,d\tau \leq max\,|x(\tau)|\, (1-0) = ||x||$. In this case, I would have that the operator is bounded. However, having $t$ which is not a constant makes me think that I can't do the same thing.
Update: I also found this link with the same problem at first but no answer.
The variable $t$ that appears in the definition of $Tx$ is the input variable. A more clear definition of $T:C[0,1] \to C[0,1]$ would read $$ (Tx)(t) = \int_0^t x(\tau)d\tau. $$ That is, $T$ maps from functions to functions and $Tx$ is itself a function. And so for any $x \in C[0,1]$ we have by definition
$$ ||Tx|| = \max_{t \in [0,1]} |\int_0^t x(\tau) d\tau|. $$ From here you should be able to prove boundedness. In particular, from your argument, we know $$ ||Tx|| \leq \max_{t\in [0,1]} \max_{\tau \in [0,1]} |{x(\tau)}|\cdot t $$ and only a little more needs to be done to finish.