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

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.

share|cite|improve this question
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:… – Jonas Meyer Jan 26 '12 at 6:30
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.

share|cite|improve this answer
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 $[0,1]$.

share|cite|improve this answer

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.