The asymptotic behavior of an absolutely continuous function with square integrable derivative

Question

Let $f : \mathbb R \to \mathbb R$ be absolutely continuous and assume that $f'\in L^2([0,1])$ and that $f(0) = 0.$ Show that the following limit exists and compute its value: $ \lim_{x \to 0} x^{-1/2} f(x).$

My approach since $f$ is absolutely continuous $f'(x)$ exists almost everywhere. And by Mean Value Theorem $f(x) -f(0) =f'(c)(x-0)$ for some $c \in (0, x)$. Which further $f'$ is in $L^2$ of $[0 1]$ which implies the square integral is finite. but from here how can i proceed?

There is no Mean Value Theorem for absolutely continuous functions. There is a Fundamental Theorem of Calculus: $f(x)-f(0)=\int_0^x f'(t),dt$. — David C. Ullrich, Nov 07 '15 at 19:41
Yea You are right. I am sorry. next what to do to find the limit? — user145993, Nov 07 '15 at 19:46
i did apply there then i got limit is 0 am i right?@John Dawkins — user145993, Nov 07 '15 at 20:53
I edited the first paragraph. Right-click on it to find the underlying LaTeX, then please edit the second paragraph. — zhw., Nov 08 '15 at 04:30

score 1 · Answer 1 · 2015-11-16T01:57:14.643

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As suggested in a comment by John Dawkins, you need the Cauchy-Schwarz inequality:

$$ f(x)^2 \le \left(\int_0^x |f'(t)|\,dt\right)^2\le \int_0^x 1\,dt \int_0^x |f'(t)|^2\,dt = x \int_0^x |f'(t)|^2\,dt $$ Here the factor $\int_0^x |f'(t)|^2\,dt$ tends to zero as $x\to 0$, and the conclusion $x^{-1}f(x)^2\to0$ follows.

edited Nov 16 '15 at 01:57

answered Nov 15 '15 at 08:44

I must be missing something but why should the factor $\int_0^x |f'(t)|^2,dt$ tends to zero as $x\to \infty$? – BigbearZzz Nov 15 '15 at 13:37
By the Absolute continuity of the Lebesgue integral – Nov 15 '15 at 16:00
I am truly sorry but I still don't get your point. Isn't the function $|f'(x)|^2$ positive so $F(x)=\int_0^x |f'(t)|^2,dt$ is an increasing function of $x$ with $F(0)=0$? – BigbearZzz Nov 16 '15 at 01:13
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But isn't that what you wrote in your answer (I'm getting very confused here)? Didn't you really mean $x\to 0$? – BigbearZzz Nov 16 '15 at 01:56
Right, was a typo. Fixed – Nov 16 '15 at 01:57

The asymptotic behavior of an absolutely continuous function with square integrable derivative

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