Supremum of the upper bound of an Integral .

Question

So I am basically drafting the proof for the convergence of a numerical SDE method and came upon an interesting problem that is more real analysis than stochastic calculus. I have to calculate
$$ \sup_{0\le t \le T} \int_0^t \left |X_n(s)-X(s)\right | ^p ds $$. I know that the integral of a non-negative function is itself a bounded and increasing function of t, so the answer should be $\int_0^T \left |X_n(s)-X(s)\right | ^p ds $ , since the integrated function does not depend on t, but I would like to hear feedback on it so as to be sure . Thank you .

Answer 1

For any $0\le t_1\le t_2$, you have :

$$\begin{align} \int_0^{t_2} \left |X_n(s)-X(s)\right | ^p ds &= \int_0^{t_1} \left |X_n(s)-X(s)\right | ^p ds +\underbrace{\int_{t_1}^{t_2} \left |X_n(s)-X(s)\right | ^p ds}_{\ge 0} \\ &\ge\int_0^{t_1} \left |X_n(s)-X(s)\right | ^p ds \end{align}$$

So it follows that $$ \sup_{0\le t \le T} \int_0^t \left |X_n(s)-X(s)\right | ^p ds = \int_0^T \left |X_n(s)-X(s)\right | ^p ds $$

As you rightfully deduced.