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Let $f:\mathbb R^n \to \mathbb C$ continuous such that $$ \int_{\mathbb{ R}^n} |f(x)|dx < + \infty $$

There is usually a claim in some Fourier Analysis lecture notes that $$ |f(x)|\to 0 \textrm{ as }|x|\to \infty $$

where can I find a good proof of this statement?

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    This is incorrect, you need stronger assumptions like uniform continuity. – 2017-01-09
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    with uniform continuity we don't need the integrability any more. – 2017-01-09
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    sure we do, consider e.g. $f(x) = 1$ or $f(x)=x$. – 2017-01-09
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    yes you are right in my mind it was the converse. saying any function vanishing at the infinity is uniformly continuous. thank – 2017-01-09
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    how do you prove this well known identity $\widehat{f'}(\xi) = c\xi \widehat{f}(\xi) $ ? – 2017-01-09
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    If you know that $f$ and $f'$ are in $L^2(\mathbb{R})$ on $\mathbb{R}$, then it is true that $f$ vanishes at $\infty$. In this case, $(f^2)'=2ff'$ is absolutely integrable, which gives a limit for $f^2$ at $\pm\infty$ using the integral. That limit can only be $0$ because $f^2$ must be integrable. – 2017-01-12
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    To see that this is false, just imagine a function whose graph looks like an evenly spaced sequence of spikes of height $1$, such that the volume beneath the $n$th spike is $1/n^2$. – 2017-02-06

5 Answers 5

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If know $\Bbb R=\bigcup\limits_{n}[n,n+1)$ along with the splitting $ [n , n+1) = [n,n+\frac{1}{2^n})\cup[ n+\frac{1}{2^n},n+1)$

Then, $$\Bbb R=\bigcup\limits_{n\ge 0} \left([n,n+\frac{1}{2^n})\cup[ n+\frac{1}{2^n},n+1)\right)$$

We set the functionn $f$ (see its construction below) \begin{equation} f(x)= \begin{cases} \underbrace{-2^{2n+2}(x-n)^2+2^{n+2}(x-n)}_{P_n}& \text{if} ~~ n\le x\le n+\frac{1}{2^n} ~~\text{for some $n\in\mathbb{N}$}\\ 0 &\text{ if}~~~j+\frac{1}{2^j}\le x\le j+1~~\text{for some $j\in\mathbb{N}$}\\ f(-x)& \text{if }~~x\lt 0. \end{cases} \end{equation}

($h =1$, $ \varepsilon_n =\frac{1}{2^n}$, $a_n = n$ and $b_n =n+\frac{1}{2^n}$)

One can check that $f$ is an even and continuous function. Further, we have $$ \int_\mathbb{R} f(x)\,dx= 2\sum_{n=0}^{\infty}\int_{n}^{n+\frac{1}{2^n}} P_n(x) dx = 2\sum_{n=0}^{\infty}-4 \left[ \frac{2^{2n}}{3}(x-n)^3-\frac{2^n}{2}(x-n)^2\right]_{n}^{n+\frac{1}{2^n}}\\ = \frac43\sum_{n=0}^{\infty} \frac{1}{2^n}= \frac{8}{3} $$ Morerover, for every $n\in\mathbb{N}$ one has $$f(n) = 0~~and ~~~f(n+\frac{1}{2^{n+1}}) = 1 $$

Therefore $$\lim_{|x| \to \infty}|f(x)|\not \to 0 $$ further, $\lim_{|x| \to \infty}|f(x)|$ does not exist.

Construction of f: The aim is to construct a family of polynomial $P_n: [a_n,b_n]\to \mathbb{R}$ of degree equals to 2 with $ P_n(a_n)=P_n(b_n)=0 ,P_n(\frac{a_n+b_n}{2}) = h$ such that $$ \int_{a_n}^{b_n} P_n(x) dx =\alpha_n ~~~\text{with } ~~\sum_{n=0}^{\infty} \alpha_n <\infty$$ where $b_n =a_n +\varepsilon_n $, $\varepsilon_n\to 0$, the sequence $(a_n)$ is increasing and the family $([a_n,b_n])_n$ is pairwise disjoint.

First step

$P_n(x) = c(x-a_n)(x-b_n) = c(x-a_n)^2-c\varepsilon_n(x-a_n) ~~$ $c\in \mathbb{R}$

and $P_n(\frac{a_n+b_n}{2}) = h$ entails $-c\left(\frac{a_n-b_n}{2}\right)^2= h$ that is $c= -\frac{4h}{\varepsilon^2_n}.$

Second Step \begin{split} \int_{a_n}^{b_n} P_n(x) dx &=&-\frac{4h}{\varepsilon^2_n} \left[ \frac{1}{3}(x-a_n)^3-\frac{\varepsilon_n}{2}(x-a_n)^2\right]_{a_n}^{b_n =a_n+\varepsilon_n}\\ &=& \frac{2h\varepsilon_n}{3} = \alpha_n \end{split}

Therefore $$ \sum_{n=0}^{\infty} \alpha_n <\infty\Longleftrightarrow\sum_{n=0}^{\infty} \varepsilon_n <\infty.$$ Whence we chose $a_n= n,2^n,.... $ and $\varepsilon_n = \frac{1}{n^s}, \frac{1}{n!},\frac{1}{2^n},\frac{1}{3^n}...$ for simplicity $a_n= n$ and $\varepsilon_n = \frac{1}{2^n}$ so that $$P_n(x) = -\frac{4h}{\varepsilon^2_n}(x-n)^2+\frac{4h}{\varepsilon_n}(x-n)$$ Last step One can easily check that the family of intervals $([n,n+\frac{1}{2^n}])_n$ is pairwise disjoint We then define the function

\begin{equation} f(x)= \begin{cases} \underbrace{-\frac{4h}{\varepsilon^2_n}(x-n)^2+\frac{4h}{\varepsilon_n}(x-n)}_{P_n}& \text{if} ~~ n\le x\le n+\frac{1}{2^n} ~~\text{for some $n\in\mathbb{N}$}\\ 0 &\text{ if}~~~j+\frac{1}{2^j}\le x\le j+1~~\text{for some $j\in\mathbb{N}$}\\ f(-x)& \text{if }~~x\lt 0. \end{cases} \end{equation} We can check that $f$ even, continuous and we have $$ \int_\mathbb{R} f(x)\,dx= 2\sum_{n=0}^{\infty}\int_{a_n}^{b_n} P_n(x) dx = 2\sum_{n=0}^{\infty} \frac{2h}{2^n3} = \frac{8h}{3} $$ Morerover, for every $n\in\mathbb{N}$ one has $f(n) = 0$ and $f(n+\frac{1}{2^{n+1}}) = h $

thus $$\lim_{|x| \to \infty}|f(x)|\not \to 0 $$ further, $\lim_{|x| \to \infty}|f(x)|$ does not exist.

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This is false, let $n=1$ and consider the function

$$f(x) = \begin{cases} 8^n(x-n) && n\le x\le n+4^{-n}, && n\ge 1 \\ 2^n+8^n(n+4^{-n}-x) && n+4^{-n}\le x\le n+2\cdot 4^{-n}, && n\ge 1 \\ 0 && o/w\end{cases}$$

Perhaps you mean that $\widehat{f}(\xi)$ vanishes at infinity, which is true.

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    This function is not continuous. – 2017-01-09
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    there is a jump at $n+4^{-n}$. But as well we have $|f(x)|\to 0$ as $|x|\to\infty$. – 2017-01-09
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    Good. But we still have $f(x)$ vanishing for $|x|> n+2.4^{-n}$ so I don't see the contradiction. How do you prove this well known identity $\widehat{f'}(\xi) = c\xi \widehat{f}(\xi) $ c? – 2017-01-09
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    @GuyFabrice thanks, yes, I clearly need more coffee: confused slope with max value. This is why I prefer parametric forms of equations. Everything is fixed now. I'm not sure I understand your third comment though. – 2017-01-09
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    @GuyFabrice the function doesn't vanish for large $x$, it is unbounded. – 2017-01-09
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    Then I have one more question. Does your parameter $n$ changing according to $x$ or it is fixed?. About my previous question, to prove the identity one might need to show that $|f(x)|\to 0$ for large $x$. Isn't it? that is why asked this QUESTION on Integrability and vanishing at infinity. – 2017-01-10
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    @GuyFabrice $n$ is not a parameter, it is an index for the piecewise function $f(x)$. Between $1$ and $1+4^{-1}$ the function increases from $0$ to $2^1$ then from there to $1+2\cdot 4^{-1}$ it decreased back to $0$. Then at $2$ it starts increasing until it hits $2^2$ at $2+4^{-2}$ and then decreases back to $0$ at $2+2\cdot 4^{-2}$ and so on. As $f(x)=2^n$ at $n+4^{-n}$ the function clearly does not go to $0$. These triangles all have height $2^n$ and base $2\cdot 4^{-n}$ so the total area of the function is the sum of the individual areas: ${1\over 2}bh={1\over 2}(2^n)(2\cdot4^{-n})=2^{-n}$ – 2017-01-10
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    Please what is the value of this function at infinity then? – 2017-02-10
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    @GuyFabrice it has no "value at infinity" it goes up and down infinitely often. Remember, the negation of "vanishes at infinity" is not "goes to a different value" it is only that it does not go to $0$, and since it goes up to $2^n$ infinitely often, it can never **stay** close to $0$. – 2017-02-10
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    I get your point now. thanks – 2017-02-10
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    @GuyFabrice my pleasure, glad I was able to resolve that for you. :) – 2017-02-10
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    In other to get the continuity from $n+2.4^{-n}$ to $n+1$ I think you have to put $+8^n$ instead of $-8^n$ in the second formula. – 2017-02-11
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    @GuyFabrice no, the slope must be negative to go back down to 0 from a positive value. – 2017-02-11
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    you are right but here the slope seems positive in both formula. with $+8^n$ inn the second one has $f(n) =f(n+2.4^{-n} ) =0$ and negative slope :) – 2017-02-11
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    @guyfabrice aha! You're right, good catch. – 2017-02-11
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Let $$ f(x) = \sum_{j=0}^\infty 2^j\operatorname{tri}(\frac{x - 2^j}{2^{2j}}), \qquad (x \in \mathbb R) $$ $f$ is continuous. (The triangles do not touch)

Graphically, $f$ consists of increasingly tall and thin spikes of triangles whose total area is finite.

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    I am sorry this function is quite fastidious to handle. by the way how do you prove: $\widehat{f'}(\xi) = c\xi\widehat{f}(\xi) $ – 2017-01-09
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For $n\in \mathbb N$ let $f(x)$ be linear on $[n-2^{-n}, n]$ and linear on $[n,n+2^{-n}]$ with $f(n-2^{-n})=f(n+2^{-n})=0$ and $f(n)=2^{n/2}.$ Let $f(x)=0$ for all other $x.$ Then $$\int_0^{\infty }f(x)dx=\sum_{n=1}^{\infty}2^{-n/2}\ne \infty.$$

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The function

$$f(x) = \left (\frac{2+\cos x}{3}\right )^{x^4}$$

is real analytic on $\mathbb R$ and satisfies $\int_\infty^\infty |f| < \infty.$ But note $f(2\pi n)= 1$ for all $n.$ I like this example, but for intuition the "triangular spike" method is probably best.

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    please how to check that this unction is integrable. I am struggling to prove it – 2017-08-21
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    I never solve this problem and I would like to – 2017-10-21
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    @GuyFsone I am using the fact that $$\int_0^{2\pi}\left (\frac{2+\cos x}{3}\right )^{n}\,dx \le \frac{C}{n^{1/2}}$$ as $n\to \infty.$ Thus for my $f$ we have $$\int_{2\pi n}^{2\pi(n+1)} f \le \frac{C'}{n^{2}} ,$$ which gives $\int_0^\infty f < \infty.$. – 2017-10-21