8
$\begingroup$

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 $f$?

$\endgroup$
  • 1
    $\begingroup$ Where do you get that idea? I never heard or read something like that before. You usually need a bunch stronger assumptions. $\endgroup$ – user251257 Jul 31 '15 at 19:01
  • 2
    $\begingroup$ @user251257 SuanDev is asking a reasonable question. Under the stated conditions, we know that UC permits the interchange of the limit and the integral. But, UC is not a necessary condition. For example, dominated convergence also permits the interchange. So, under what added conditions does UC follow from the equality of the limit of the integral and the integral of the limit? $\endgroup$ – Mark Viola Aug 1 '15 at 2:27
  • $\begingroup$ @Dr.MV : Thank you ; that is exactly the point form where I am asking the question . $\endgroup$ – user228168 Aug 1 '15 at 6:03
  • $\begingroup$ @SaunDev You're welcome ... my pleasure. $\endgroup$ – Mark Viola Aug 1 '15 at 6:25
  • $\begingroup$ I didn't want it to sound rude. I was just curious. This question and its answer might give you an idea how much stronger UC is: math.stackexchange.com/q/1353844/251257 $\endgroup$ – user251257 Aug 1 '15 at 8:25
1
$\begingroup$

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 $[a, b]$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy