When does interchangibility of limit and Riemann integral imply uniform convergence?

Question

Let $\{f_n\}$ be a sequence of real-valued functions defined on an interval $[a,b]$ such that each $f_n$ is Riemann integrable, $\{f_n\}$ converges point-wise to $f$, $f$ is Riemann integrable and $\lim_{n \to \infty} \int_a^b f_n$ exists and equals $\int_a^b f$; then under what additional condition(s) can we conclude that $\{f_n\} $ converges uniformly to $f$?

Answer 1

If $g_n = f_n - f$ is continuous on $[a, b]$ then $f_n$ converges uniformly to $f$ on $[a, b]$ if and only if

$$\lim_{n \to \infty} \sup_{a\leq a' < b' \leq b}\frac{1}{b' - a'}\int_{b'}^{a'}\left| f_n(x) - f(x) \right|dx = 0$$

(Note: last one includes $\lim_{n \to \infty}\int_{b}^{a}f_n(x)dx = \int_{b}^{a}f(x)dx$)

In order to prove first part let's assume $f_n$ converges uniformly on $[a, b]$. Then given $\epsilon > 0$ exists $n' \in \mathbb{N}$ such that

$$\sup_{x \in [a, b]}\left| g_n(x) \right|< \epsilon $$

when $n > n'$.

Which implies

$$\int_{b'}^{a'}\left| g_n(x)\right|dx < \int_{b'}^{a'}\epsilon dx = (b' - a')\epsilon $$ when $a \leq a' < b' \leq b$.

Now $$ \sup_{a\leq a' < b' \leq b}\frac{1}{b' - a'}\int_{b'}^{a'}\left| g_n(x) \right|dx < \frac{b' - a'}{b' - a'} \epsilon = \epsilon$$

Because $\epsilon$ was arbitrary we have shown $$\lim_{n \to \infty} \sup_{a\leq a' < b' \leq b}\frac{1}{b' - a'}\int_{b'}^{a'}\left| f_n(x) - f(x) \right|dx = 0$$

To prove second part let's assume $\lim_{n \to \infty} \sup_{a\leq a' < b' \leq b}\frac{1}{b' - a'}\int_{b'}^{a'}\left| f_n(x) - f(x) \right|dx = 0$.

Now given $\epsilon > 0$ exists $n' \in \mathbb{N}$ so that when $n > n'$

$$ \sup_{a\leq a' < b' \leq b}\frac{1}{b' - a'}\int_{b'}^{a'}\left| g_n(x) \right|dx < \frac{1}{2}\epsilon$$

Because $g_n = f_n - f$ is continuous for any $x \in [a, b]$ exists $[a', b'] \subset [a, b]$ such that $\left| g_n(t) - g_n(x) \right| < \frac{1}{2}\epsilon$ and $x \in [a', b']$ when $t \in [a', b']$.

Using mean value theorem we get

$$g_n(x) - \frac{1}{2}\epsilon <\frac{1}{b' - a'} \int_{a'}^{b'}g_n(t) dt < g_n(x) + \frac{1}{2}\epsilon$$

$$\frac{1}{b' - a'} \int_{a'}^{b'}g_n(t) dt - \frac{1}{2}\epsilon < g_n(x)<\frac{1}{b' - a'} \int_{a'}^{b'}g_n(t) dt + \frac{1}{2}\epsilon$$

$$- \frac{1}{2}\epsilon - \frac{1}{2}\epsilon < g_n(x)< \frac{1}{2}\epsilon + \frac{1}{2}\epsilon$$

$$-\epsilon < g_n(x)< \epsilon$$ Because this is for any $\epsilon > 0$ or $x \in [a, b]$ we get $\sup_{x \in [a, b]}\left| g_n(x) \right|< \epsilon $ and $\lim_{n \to \infty} \sup_{x \in [a, b]}\left| g_n(x) \right| = 0$. Thus $f_n$ converges uniformly to $f$ on $[a, b]$.