Uniform Convergence of a time-dependent Riemann Sum

Question

Suppose $g:[a,b]\times[0,\infty)\to \mathbb{R}$ is a smooth function. For each $n\in\mathbb{N}$, let $P_n$ be an equally-spaced partition of $[a,b]$ of size $n$. Are there any conditions under which the sequence of Riemann sums $$ f_n(t) := \sum_{x_j\in P_n} g(x_j,t) \Delta x_j $$ converge uniformly to the integral $$ f(t) := \int_a^b g(x,t) dx $$ as $n\to\infty$?

Answer 1

A sufficient condition is that $g$ is uniformly continuous on $[a,b]\times[0,\infty)$. This would imply for any $\epsilon > 0$ there exists $\delta(\epsilon) > 0$ such that for all $x,y \in [a,b]$ and for all $s,t \in [0,\infty)$ such that $|x-y| < \delta(\epsilon)$ and $|s-t| < \delta(\epsilon)$ we have $|g(x,t) - g(y,s)| < \epsilon$.

In particular, for all $t \in [0,\infty)$ and $|x-y| < \delta(\epsilon/(b-a))$ we have $|g(x,t) - g(y,t)| < \epsilon/(b-a)$.

For any partition $P = (x_0,x_1,\ldots,x_n)$ of $[a,b]$ it follows that

$$\left|\sum_{k=1}^n g(\xi_k,t)(x_k - x_{k-1}) - \int_a^b g(x,t) \, dx \right| = \left|\sum_{k=1}^n \int_{x_{k-1}}^{x_k} [g(\xi_k,t)-g(x,t)] \, dx \right| \\ \leqslant \sum_{k=1}^n \int_{x_{k-1}}^{x_k} |g(\xi_k,t)-g(x,t)| \, dx, $$

where $\xi_k$ could be any intermediate point in $[x_{k-1}, x_k]$.

Hence, if $\|P\| = \max_{1\leqslant k \leqslant n}(x_k - x_{k-1}) < \delta(\epsilon/$(b-a)) we have $|g(\xi_k,t) - g(x,t)| < \epsilon/(b-a)$ for all $t \in [o,\infty)$, and

$$\left|\sum_{k=1}^n g(\xi_k,t)(x_k - x_{k-1}) - \int_a^b g(x,t) \, dx \right| \leqslant \sum_{k=1}^n \int_{x_{k-1}}^{x_k} \frac{\epsilon}{b-a}\, dx = \epsilon, $$

Therefore, the convergence of Riemann sums to the integral is uniform when $g$ s uniformly continuous.