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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

  • 0
    A related question: http://math.stackexchange.com/questions/82862/f-indefinite-integral-implies-f-is-absolutely-continuous2012-01-26

2 Answers 2

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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.

  • 0
    There was a [related question](http://math.stackexchange.com/questions/10764) last November regarding the Lipschitz case.2011-05-21
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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 $[0,1]$.