Let $f:\mathbb R→\mathbb R$ s.t. $\int_{-\infty}^\infty |f(x)| \, dx<\infty$ and $F:\mathbb R→\mathbb R$ defined as $F(x)=\int_{-\infty}^x f(t) \, dt$. Then how to prove that $F(x)$ is uniformly continuous?

-I have got no idea how to prove it. However, I know the definition of uniform continuity and the sufficient condition of bounded derivative to prove it. As $f(x)$ may be unbounded too, which suggests that $F'(x)=f(x)$ may not bounded in some case thereby restricting me to use the sufficient condition of uniform continuity. Kindly help.


1 Answer 1


It suffices to show that for any $\epsilon>0$ there exists $\delta>0$ such that $$x<y, |x-y| \leq \delta \implies \int_x^y |f(t)| \, dt < \epsilon. \tag{1}$$

Hints to prove $(1)$: Fix $\epsilon>0$.

  1. We have $$\int_x^y |f(t)| \, dt = \int_{[x,y] \cap \{|f| \geq R\}} |f(t)| \, dt + \int_{[x,y] \cap \{|f|<R\}} |f(t)| \, dt =: I_1+I_2$$ for any fixed constant $R>0$ (which will be chosen in the next step).
  2. Since $f$ is integrable, we can choose $R>0$ sufficiently large such that $$I_1 \leq \int_{|f| \geq R} |f(t)| \, dt \leq \epsilon.$$ Note that $R$ does not depend on $x$ and $y$.
  3. As $$I_2 \leq R |x-y|,$$ we get $$\int_x^y |f(t)| \, dt \leq R |x-y| + \epsilon.$$
  4. Conclude.
  • 2
    $\begingroup$ For once, I think you missed the "good" argument, which would be to (re)prove that every continuous function on $\mathbb R$ with limits at $\pm\infty$ is uniformly continuous. $\endgroup$
    – Did
    Sep 27, 2016 at 6:03
  • $\begingroup$ @saz....Will it be possible to prove that $F(x)$ is Lipschitz continuous? $\endgroup$ Sep 27, 2016 at 6:09
  • 1
    $\begingroup$ From your equation (1), $|f(x)-f(y)|=\int_x^y |f(t)|dt=|f(c)||x-y|\leq |x-y|$, where $c\in (x,y)$; hence $F(x)$ is Lipschitz continuous. Please correct me if I am wrong. $\endgroup$ Sep 27, 2016 at 6:14
  • $\begingroup$ @Did I agree with you that your reasoning is somewhat nicer ... you could just write another answer to the question... $\endgroup$
    – saz
    Sep 27, 2016 at 6:18
  • 1
    $\begingroup$ @mathlover I suppose you mean $|F(x)-F(y)| = \dots$. However, your reasoning works only if $f$ is bounded. If $f$ is not bounded, then $F$ is in general not Lipschitz continuous (consider e.g. $f(x) = 1_{(0,\infty)}(x) x^{-1/2}$). $\endgroup$
    – saz
    Sep 27, 2016 at 6:21

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .