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I've been trying to show that a function $f$ on the real interval $[a,b]$ which satisfies

$$ f(x)=f(a)+\int_a^xf'(s)\,ds\qquad\text{($f'$ defined almost everywhere)} $$

must be uniformly continuous on $[a,b]$.

Since the condition above is equivalent to absolute continuity I know that I could show what I need by means of the proof that absolute continuity - from its fundamental definition - implies uniform continuity: I have seen a proof of that. However, I would like to show the above without involving another form of continuity in the process.

I know that - since what I have stated is also equivalent to there existing any integrable function in place of $f'$ -the proof should not involve the properties of the derivative. However, in establishing a bound I get only as far as

$$ \left|f(y)-f(x)\right|=\left|\int_x^yf'(s)ds\right|\leq\int_x^y\left|f'(s)\right|\,ds $$

Is it possible to show that an integrable first derivative (or indeed any integrable function) is bounded in sup norm?

Thank you.
Marko

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3  
It seems to me this should be simple. If $f'$ is integrable then $f$ must be continuous (dominated convergence argument), and a continuous function on a compact set (such as $[a,b]$) is always uniformly continuous. –  Nate Eldredge May 20 '11 at 22:31
    
Thanks very much guys. I guess I had to choose one of the answers but they're all equal almost everywhere. And I got five points for a silly question, so everyone's a winner. :) –  Josef K. May 23 '11 at 22:19
    
A related question: math.stackexchange.com/questions/82862/… –  Jonas Meyer Jan 26 '12 at 6:30

2 Answers 2

up vote 3 down vote accepted

As you mentioned, $f$ is absolutely continuous, and showing this isn't really harder than showing uniform continuity directly.

If $g$ is integrable and $\varepsilon>0$ is given, there is a $\delta>0$ such that $m(A)<\delta$ implies $\int_A|g|<\varepsilon$. To see this, you could for instance first take $h$ bounded by $M>0$ such that $\int_a^b|g-h|<\frac{\varepsilon}{2}$, and then take $\delta = \frac{\varepsilon}{2M}$.

Once you have this, you have $|\int_x^y g|<\varepsilon$ whenever $|x-y|<\delta$. And as mentioned, this extends to showing absolute continuity. Boundedness of $f'$ would imply the stronger condition of Lipschitz continuity.

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After posting my correction I see that I'm definitely ripe for bed: your post is probably a hundred times more to the point than mine. +1 of course. –  t.b. May 20 '11 at 23:45
    
@Theo: For one thing, it is more concise largely because I left out mention of a method to obtain $h$, whereas you gave details of one such method along with other details which I omitted. For another thing, I did benefit from it not being late in my timezone:) +1 to yours, too. –  Jonas Meyer May 21 '11 at 1:51
    
Thanks for the correction of the typos in my post! Still, I like your answer much more than mine, even after sleeping. Here's one little addendum: boundedness of $f'$ is equivalent to Lipschitz continuity. This, or rather its generalization to functions $f:U \to \mathbb{R}^n$ defined on an open subset $U$ of $\mathbb{R}^m$ is called Rademacher's theorem, as you certainly know. –  t.b. May 21 '11 at 6:49
    
@Theo: I was vaguely aware of that generalization, but didn't know its name, so thanks very much for the link (and compliment)! But yes, the 1-dimensional case is straightforward once you have the absolute continuity theory in hand. If $f$ is Lipschitz, then by absolute continuity $f'$ exists a.e. (even BV is enough for that part) and $f(x)=f(a)+\int_a^xf'$. Boundedness of $f'$ by the Lipschitz constant of $f$ everywhere $f'$ exists is immediate from the definitions, and Lipschitz continuity of $x\mapsto \int_a^x g$ for bounded measurable $g$ is pretty much as immediate. –  Jonas Meyer May 21 '11 at 7:15
    
There was a related question last November regarding the Lipschitz case. –  Jonas Meyer May 21 '11 at 7:19

We assume that $\|f'\|_{L^1} = \int_{a}^{b} |f'|\,dt \lt \infty$. Let $A_{n} = \{x \,:\,|f'(x)| \leq n\}$. Put $g_{n} = [A_{n}] f'$, where $[A_n]$ denotes the characteristic function of $A_n$. Then we have $g_{n} \to f'$ almost everywhere, and, as Nate points out in his comment, dominated convergence implies that $\int_{a}^{b} |g_n - f'|\,dt \to 0$ as $n \to \infty$ (the integrand is bounded by the integrable function $2|f'|$).

Now, given $\varepsilon \gt 0$, choose $n$ so large that $\int_{a}^{b} |g_n - f'|\,dt \lt \varepsilon /2$. As $|g_{n}|$ is bounded by $n$, we have that $|\int_{x}^{y} g_{n}(t)\,dt| \leq n|y-x|$. Thus, $$\left\vert \int_{x}^{y} f'(t)\,dt\right\vert \leq \left\vert\int_{x}^{y} |f'-g_n|\,dt\right\vert + \left\vert \int_{y}^{x} |g_n(t)|\,dt\right\vert \leq \varepsilon/2 + n \cdot |y-x|$$ and for $\delta = \frac{\varepsilon}{2n}$ we get for all $x,y$ with $|y-x| \lt \delta$ that $$|f(y) - f(x)| = \left\vert \int_{x}^{y} f'(t)\,dt \right\vert \leq \varepsilon/2 + \delta n \lt \varepsilon$$ which is the very definition of uniform continuity of $f$.

In fact, we get the even more general estimate that for $\mu(E) \lt \delta$ we have $\int_{E} |f'|\,dt \lt \varepsilon$. But that's exactly absolute continuity of $f$.

It is of course not true that an integrable first derivative is bounded in the sup-norm. For instance, for $f(x) = \sqrt{x}$ we have $f'(x) = \frac{1}{2\sqrt{x}}$ which is not bounded but integrable on $[0,1]$.

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