Let $f,f'\in L^1(\mathbb R)$. Prove that $$\lim_{x\to \infty }f(x)=0.$$

First of all, is $f'$ defined a.e. ? Because there are no assumption on the fact that $f$ is derivable. So, is $$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$ for almost every $x$ ?

My attempt for the statement : Suppose WLOG $f'>0$. I'm not sure if it's true, but I would say that : Let $\varepsilon >0$. Since $f'$ is $L^1$, there is a ball $[a,b]$ s.t. $$\int_{]-\infty ,a[\cup]b,+\infty [}f'(x)dx<\varepsilon .$$ In particular, if $x>y>b$, then $$f(x)-f(y)=\int_y^x f'(t)dt<\varepsilon,$$ but I just can conclude that $\lim_{x\to \infty }f(x)-f(y)<\varepsilon .$

  • $\begingroup$ For the first question, why not? It's written "$f' \in L^1(\Bbb{R})$" in the title. If $f'$ is not defined a.e., this condition would be meaningless. It suffices to show the case for $f\ge0$. Try a proof by contradiction. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 May 18 at 8:34
  • $\begingroup$ @GNUSupporter8964民主女神地下教會: Thanks but how do you think $f'$ is defined ? In the weak sense or as I defined in my post ? $\endgroup$ – user659895 May 18 at 8:40
  • $\begingroup$ @user659895 Do you know that there exist continuous strictly increasing functions $f$ with $f'=0$ almost everywhere. Fundamental Theorem of Calculus (FTC) is proved under the assumption that $f$ has a continuous derivative and you cannot make such assumptions in this question. So use of FTC is not at all admissible here. $\endgroup$ – Kavi Rama Murthy May 18 at 23:15

I think the question is not properly posed. Normally the statement is interpreted as follows: if $f$ is integrable and differentiable almost everywhere with $f'$ also integrable then $f(x) \to 0$ as $ x \to \infty$. This statement is false. $f(x)=1$ when $x$ is an integer and $0$ otherwise gives a counterexample. Some additional assumptions on $f$ are necessary to prove that $f(x) \to 0$ as $ x \to \infty$.

  • $\begingroup$ but $f=1$ is not integrable... $\endgroup$ – user659895 May 18 at 12:04
  • $\begingroup$ $f=1$ only at integer points which form a set of measure $0$. @user659895 $\endgroup$ – Kavi Rama Murthy May 18 at 12:05
  • $\begingroup$ Strange because to prove that the fourier transform of the derivative is $\mathcal F(f')(\alpha )=-2i\pi \alpha \mathcal F(f)(\alpha )$, in my exercise the only assumption are $f,f'\in L^1$ $\endgroup$ – user659895 May 18 at 12:07
  • $\begingroup$ In such results it is generally assumed that $f$ is an absolutely continuous function such that $f$ and $f'$ are integrable. $\endgroup$ – Kavi Rama Murthy May 18 at 12:09
  • $\begingroup$ It appears that the downvoter doesn't care about Mathematical rigor. $\endgroup$ – Kavi Rama Murthy May 18 at 12:19

The idea is to show that $f(x)-f(0)=\int_0^x f'$ hence $f$ has a limit in $\infty$, which must be $0$ because $f\in L^1$.


On the one hand, observe that exists $l = \lim_{x \to \infty} f(x)$. It is because for each $x \in \mathbb{R}$ $$ f(x) - f(a) = \int_{(a , x)} f' \quad \Longrightarrow \quad \lim_{x \to \infty} f(x) = f(a) + \int_{(a , \infty)} f' \in \mathbb{R}\mbox{,} $$ as $f' \in L^1(\mathbb{R})$. On the other hand, if $l \neq 0$, then there exist $\varepsilon , x_0 \in (0 , \infty)$ such that $|f(x)| \geq \varepsilon$ for all $x \in \mathbb{R}$ with $x \geq x_0$. Therefore $$ \int_{\mathbb{R}} |f| \geq \int_{[x_0 , \infty)} |f| \geq \varepsilon \int_{[x_0 , \infty)} 1 = \infty\mbox{,} $$ which is a contradiction, as $f \in L^1(\mathbb{R})$.

  • $\begingroup$ You are making some strong assumptions. The hypothesis does not guarantee that the Fundamental Theorem of Calculus is applicable. $\endgroup$ – Kavi Rama Murthy May 18 at 11:54
  • $\begingroup$ Hypothesis guarantee that the FTC is applicable because $f$ is continuous on $\mathbb{R}$, you can remember the FTC's hypothesis here: en.wikipedia.org/wiki/Fundamental_theorem_of_calculus $\endgroup$ – joseabp91 May 18 at 12:02
  • $\begingroup$ It is not given that $f$ is continuous and it is not given that $f$ is differentiabel at every point. The statement, as it stands, is false as I have pointed out in my answer. $\endgroup$ – Kavi Rama Murthy May 18 at 12:04
  • $\begingroup$ I think that you are wrong: if $f \in L^1(X)$, normally we suppose that $f : X \to \mathbb{R}$. $\endgroup$ – joseabp91 May 18 at 12:11
  • $\begingroup$ In analysis we often talk of $f'$ even when $f$ need not be everywhere-differentiable, only requiring that $f$ be differentiable almost-everywhere instead. Under this interpretation, FToC need not hold. Even if $f$ is everywhere differentiable and $f'$ is integrable, it is a quite non-trivial result that FToC holds for this case, which is often skipped in many analysis textbook. (Rudin's RCA is a notable exception, and he proves this fact using Vitali–Carathéodory theorem.) $\endgroup$ – Sangchul Lee May 19 at 7:00

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