$X={f\in C(0,1): f^{\prime \prime }(t)<0}$.
$T:X\mapsto X$ is defined by:
$Tf(t) = at+ (1-t)\int_{0}^{t}sf(s)ds + t\int_{t}^{1}(1-s)f(s)ds$
and it is known that $\int_{0}^{1}s(1-s)f(s)ds < \infty$.
So that I have:
$|\frac{d}{dt}Tf(t)| < a + g(t)$.
where: $g(t) = \int_{0}^{t}sf(s)ds + \int_{t}^{1}(1-s)f(s)ds$
My instructor makes the following claim:
If we can show that g(t) is in $L^1$, then we have shown that $\frac{d}{dt}Tf(t)$ is uniformly bounded.
I have showed that indeed $g(t) \in L^1$.
But I do not know how this implies that the derivative is uniformly bounded