Doubt about inequalities and integrals - on proving an operator is well defined. Exercise. Consider the operator $T\colon \mathcal C[0,1] \to \mathcal C[0,1]$ defined by
$$ (Tf)(x) = \int_0^x f(t) \,  dt \quad \forall f \in \mathcal C[0,1].$$
Show that T is well-defined.
My solution. To show that $T$ is well-defined, the crucial part is to show that $Tf \in \mathcal C[0,1],$ for all $f \in \mathcal C[0,1].$ To do so, one must show that $Tf$ is continuous in the closed interval $[0,1].$
Thus, fix a function $f \in \mathcal C[0,1],$ a point $c \in [0,1]$ and a value $\varepsilon >0,$ all arbitrarily.  Then,
$$ | (Tf)(x) - (Tf)(c) | = \Bigg| \int_0^x f(t) \, dt - \int_0^c f(t) \, dt \Bigg| = \Bigg| \int_c^x f(t) \, dt \Bigg| \color{red}{\leqslant \int_c^x |f(t)| \, dt }.$$
Now, since $f$ is continuous in a compact interval, $f$ is also bounded and thus $|f(t)| \leqslant M,$ for some $M>0.$ Thus, based on the inequality we just found,
$$ |(Tf)(x) - (Tf)(c)| \color{red}{\leqslant \int_c^x |f(t)|\, dt}\color{purple}{ \leqslant \int_c^x M dt} = M(x-c) \leqslant M|x-c|.$$
With this, if we take $|x-c| < {\displaystyle \frac{\varepsilon}{M}}$ it follows that
$$ \forall \varepsilon > 0, \exists \delta = \delta(\varepsilon) > 0\colon |x-c| < \delta \Rightarrow |(Tf)(x) - (Tf)(c)| < \varepsilon, \quad \forall c \in [0,1].$$
proving the continuity of $Tf$ in $[0,1].$
My concerns. My only problem with my solution is applying integration to inequalities. I know the following general results hold:

Theorem. Let $f$ be a real/complex function continuous in the closed interval $[a,b].$ Then,
$$ \Bigg| \int_a^b f(t) \, dt \Bigg| \leqslant \int_a^b |f(t)| \, dt. $$

I have used this result in the steps I colored in $\color{red}{red}$ but I am not sure about one thing: I am certainly not guaranteed that $x>c$. Does this inequality also hold if $x \leqslant c?$ If so,why?
The inequation in $\color{purple}{purple}$ represents the same exact problem, because we know that $|f(t)| \leqslant M$ and I can apply integrals in both sides of the inequality if $x>c$, but what if $x \leqslant c?$
Thanks for any help in advance.
 A: You have correctly shown that
$$ \tag{$*$}
| (Tf)(x) - (Tf)(c) |  \le M |x-c|
$$
if $0 \le c \le x \le 1$.
But if $0 \le x \le c \le 1$ then
$$
| (Tf)(x) - (Tf)(c) | = | (Tf)(c) - (Tf)(x) | \underset{(*)}{\le }
 M |c-x| \le M |x-c|
$$
so that $(*)$ holds as well.
Therefore $(*)$ holds for all $x, c \in [0, 1]$, which proves that $Tf$ is (Lipschitz) continuous.
In other words: The inequality $(*)$ is symmetric in $x$ and $c$, therefore it suffices to prove it for the case $c \le x$.
Remark: Even more is true: The “Fundamental theorem of calculus“ states that $Tf$ is differentiable if $f$ is continuous.
A: That inequality is actually almost always false when $b<a$, because then $\int_a^b|f(t)|\,\mathrm dt\leqslant 0$, whereas $|\int_a^bf(t)\,\mathrm dt|\geqslant0$.
But you have$$-|b-a|M\leqslant\int_a^bf(t)\,\mathrm dt\leqslant|b-a|M,$$since you always have $-M\leqslant f(t)\leqslant M$, and therefore$$\left|\int_a^bf(t)\,\mathrm dt\right|\leqslant|b-a|M.$$So, if $|x-c|<\frac\varepsilon M$, then $\left|\int_a^bf(t)\,\mathrm dt\right|<\varepsilon$.
