Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The following statement should be true, I think, but I'm having a hell of a time trying to prove it:

Let $f_n$ be a $C^1$ function on $[0,a]$, satisfying $f_n = 1$ on $[1/n,a]$ and $0\le f_n \le 1$ and $f_n(0)=0$. Let $\phi$ be a continuous function on $[0,a].$

I want to say that

$\lim_{n\to\infty} \int_0^a f_n'(x) \phi(x) dx = \phi(0)$.

If $\phi$ is $C^1$, I can just do integration by parts to prove this. But I'm not sure what to do if $\phi$ is just continuous.

This is what I have so far. Using the mean value theorem, we have

$\int_0^a f_n'(x)\phi(x) dx = \int_0^{1/n} f_n'(x)\phi(x) dx= \frac{1}{n}f_n'(c_n) \phi(c_n)$

where $0<c_n<1/n$.

I think I need to use the mean value theorem on $f_n$ and continuity of $f_n'$ to show that $\frac{1}{n}f_n'(c_n)$ goes to 1 as $n\to\infty$, but I can't seem to figure out how to do this... can someone throw me a hint?

share|cite|improve this question
Although I prefer "helluva", (+1) for the first line :) – The Chaz 2.0 Jul 28 '11 at 17:50
There must be something missing here. $f_n\equiv1$ satisfies the conditions, and in that case the integral vanishes, independent of $\phi$. (Also the limit doesn't say what limit is being taken -- presumably $n\to\infty$?) – joriki Jul 28 '11 at 17:50
@joriki - Yeah, I forgot $f_n(0)=0$. – Braindead Jul 28 '11 at 17:56
@joriki - Yeah, I also forgot $n\to\infty$ :p – Braindead Jul 28 '11 at 17:58
@The Chaz - No wonder... I was trying to prove something that was false :S – Braindead Jul 28 '11 at 23:00
up vote 4 down vote accepted

For a concrete counterexample, consider the following.

Define $f_n$ on $[0,1/n]$ by $$ f_n (x) = \cos ^2 \bigg(\frac{\pi }{x}\bigg), \;\; x \in [\alpha_n,1/n], $$ and $$ f_n (x) = 0, \;\; x \in [0,\alpha_n], $$ where $\alpha_n$ is very small compared to $1/n$, and $f_n(\alpha_n)=0$. Note that thus $0 \leq f_n \leq 1$, $f_n (0) = 0$, and $f_n (1/n) = 1$, as required. Then, restricted to the interval $[\alpha_n,1/n]$, $$ f'_n (x) = 2\cos \bigg(\frac{\pi }{x}\bigg)\sin \bigg(\frac{\pi }{x}\bigg)\frac{\pi }{{x^2 }} = \sin \bigg(\frac{{2\pi }}{x}\bigg)\frac{\pi }{{x^2 }}, $$ and, restricted to the interval $[0,\alpha_n]$, $$ f'_n (x) = 0. $$ Note that the issue of smoothness of $f_n$ and $f'_n$ at the endpoints $\alpha_n$ and $1/n$ plays no role for our purposes. Now, define $\phi $ by $$ \phi(x) = x\sin \bigg(\frac{{2\pi }}{x}\bigg) ,\;\; x \in (0,a], $$ and $$ \phi(0) = 0. $$ Note that thus $\phi$ is continuous on $[0,a]$. Now, $$ \int_0^{1/n} {f'_n (x)\phi (x)\,dx} = \int_{\alpha _n }^{1/n} {f'_n (x)\phi (x)\,dx} = \pi \int_{\alpha _n }^{1/n} {\frac{{\sin ^2 (\frac{{2\pi }}{x})}}{x} \,dx} . $$ A change of variable $x \mapsto 1/x$ then gives $$ \int_0^{1/n} {f'_n (x)\phi (x)\,dx} = \pi \int_n^{1/\alpha _n } {\frac{{\sin ^2 (2\pi x)}}{x}\,dx} . $$ The right-hand side tends to $\infty$ if, for example, $1/\alpha_n > n^2$; then, in particular, $$ \int_0^{1/n} {f'_n (x)\phi (x)\,dx} \not \to \phi (0). $$

EDIT: Particular choice of $\alpha_n$ is not essential here. This follows from the fact that, given any $M > 0$, $\int_M^x {\frac{{\sin ^2 (t)}}{t}\,dt} \to \infty$ as $x \to \infty$.

EDIT 2: In fact, $f_n \in C^1 [0,a]$. To show this, it suffices to show that $\lim _{x \downarrow \alpha _n } f'_n (x) = 0$ and $\lim _{x \uparrow 1/n} f'_n (x) = 0$ (this would imply that $f'_n (\alpha_n) = 0$ and $f'_n(1/n)=0$); indeed, $$ \mathop {\lim }\limits_{x \downarrow \alpha _n } f'_n (x) = f'_n (\alpha _n + ) = 2\cos \bigg(\frac{\pi }{{\alpha _n }}\bigg)\sin \bigg(\frac{\pi }{{\alpha _n }}\bigg)\frac{\pi }{{\alpha _n^2 }} = 0 $$ (since, by definition, $ \cos ^2 (\frac{\pi }{{\alpha _n }}) = f_n (\alpha_n) = 0$) and $$ \mathop {\lim }\limits_{x \uparrow 1/n } f'_n (x) = f'_n \bigg(\frac{1}{n} - \bigg) = 2\cos \bigg(\frac{\pi }{{1/n }}\bigg)\sin \bigg(\frac{\pi }{{1/n }}\bigg)\frac{\pi }{{1/n^2 }} = 0 $$ (since $\sin(n \pi)=0$).

share|cite|improve this answer
I put the integral into mathematica with $\alpha_n = 2/(2(n+k)+1)$, set $n>0$ integer, $k\ge0$ integer, and then took the limit as $n\to\infty$, and I got 0... – Braindead Jul 28 '11 at 22:38
Nice; this is the kind of example I was thinking of. – joriki Jul 28 '11 at 22:40
@Braindead: Yes, that doesn't satisfy $1/\alpha_n > n^2$. – joriki Jul 28 '11 at 22:40
@joriki: Indeed! I skipped an important part. – Braindead Jul 28 '11 at 22:48

If $\phi$ is not of bounded variation (and thus not $C^1$), this may not hold. If $f_n$ is monotonic, you can directly use the continuity of $\phi$:

$$ \begin{eqnarray} \lim_{n\to\infty}\left|\int_0^af_n'(x)\phi(x)\mathrm dx-\phi(0)\right| &=& \lim_{n\to\infty}\left|\int_0^{1/n}f_n'(x)\phi(x)\mathrm dx-\phi(0)\right| \\ &=& \lim_{n\to\infty}\left|\int_0^{1/n}f_n'(x)\phi(x)\mathrm dx-\int_0^{1/n}f_n'(x)\phi(0)\mathrm dx\right| \\ &=& \lim_{n\to\infty}\left|\int_0^{1/n}f_n'(x)(\phi(x)-\phi(0))\mathrm dx\right| \\ &\le& \lim_{n\to\infty}\int_0^{1/n}\left|f_n'(x)\right|\left|\phi(x)-\phi(0)\right|\mathrm dx \\ &\le& \lim_{n\to\infty}\int_0^{1/n}\left|f_n'(x)\right|\epsilon(n)\mathrm dx \\ &=& \lim_{n\to\infty}\int_0^{1/n}f_n'(x)\epsilon(n)\mathrm dx \\ &=& \lim_{n\to\infty}\epsilon(n) \\ &=& 0\;. \end{eqnarray} $$

However, if $f_n$ is not monotonic and $\phi$ is not of bounded variation, you can let $f_n$ oscillate enough in sync with oscillations in $\phi$ that the limit may not exist.

share|cite|improve this answer
Tell me what is wrong with the following argument for $\phi$ being C^1: $\int_0^{1/n} f'_n(x)\phi(x) = f_n(x)\phi(x)|_0^{1/n} - \int_0^{1/n} f_n(x) \phi'(x)dx$. Letting $n\to\infty$, the rhs goes to $\phi(0)$. – Braindead Jul 28 '11 at 19:14
@Braindead: Sorry, you're right; I had forgotten to take into account $0\le f_n\le1$. – joriki Jul 28 '11 at 22:04

Your argument for $C^1$ $\phi$ is fine. In fact it can be extended to work for continuous bounded variation $\phi$ if you know a bit about the Riemann-Stieltjes integral since this has a more powerful integration by parts available (not requiring the functions to be differentiable). Specifically:

$$ \int_0^{1/n} \phi f_n'dx = \int_0^{1/n} \phi df_n = f_n(1/n)\phi(1/n) - f_n(0)\phi(0) - \int_0^{1/n} f_n d \phi = \phi(1/n) - \int_0^{1/n} f_n d \phi$$

Clearly $\phi(1/n) \to \phi (0)$. The last integral is bounded in magnitude by $F_n \cdot V_0^{1/n} \phi$ where $F_n = \max_{x \in [0,1/n]} f_n(x)$ and $V_0^{1/n} \phi$ is the total variation of $\phi$ on $[0,1/n]$. Assuming $\phi$ is bounded variation on $[0,\epsilon)$ for some $\epsilon >0$, both of these quantities go to zero as $n \to \infty$ so we are done.

Added: The example I originally gave to demonstrate that this can fail for non-BV $\phi$ was wrong. Here is another one. Let $\phi(x) = x \sin(X)$ for $x \neq 0$ and let $\phi(0) = 0$. Joriki's idea is the right one. Any of the integrals $\int_0^{1/n} f_n' \phi$ can be made as large as desired by making $f_n'$ oscillate in tandem with $\phi$. Fix some $n$ and choose $p \in (0,1/n)$ such that $\phi(p) = 0$. We put $f_n'(x) = A \phi^+ - B \phi^-$ if $x \in [0,p]$ and $f_n'(x) = 0$ for $x \geq p$. Here $\phi^+,\phi^-$ are the positive and negative parts of $\phi$ defined by $\phi = \phi^+ - \phi^-$ and $|\phi| = \phi^+ + \phi^-$. $A,B$ are positive constants chosen so that $f_n(x) = \int_0^x f_n'$ has $f_n(1) =1$. The thing to notice is that $A,B$ can be chosen arbitrarily large subject to this constraint and that this means

$$ \int_0^{1/n} f_n' \phi = A \int_0^p (\phi^+)^2 + B \int_0^p (\phi^-)^2 $$

can also be made arbitrarily large.

share|cite|improve this answer
@Theo: I was assuming continuity yes. I'll make that more clear. – Mike F Jul 28 '11 at 20:05
To have $f_n(1/n) = 1$, you need an extra factor of $n$. – Braindead Jul 28 '11 at 20:11
I think you can show that the integral in the example you gave is less than 2/n. – Braindead Jul 28 '11 at 20:17
@Braindead: Yeah I came to a similar conclusion. It doesn't work. I'll take it out for now. – Mike F Jul 28 '11 at 20:20
@Braindead: Okay I added a new example. – Mike F Jul 28 '11 at 21:30

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.