I was wondering whether the following statement is true and, if so, how it can be shown:

If $ f \in L^{1}_{Loc}(\mathbb{R}^n) $ and if for all compactly supported continuous functions $ g: \mathbb{R}^n \to \mathbb{C} $ we have that the Lebesgue integral of $ f $ multiplied by $ g $ equals zero, i.e. $$ \int_{\mathbb{R}^n} f(x)g(x) \mathrm{d}x = 0 , $$ then $ f(x) = 0 $ almost everywhere.

I would be very grateful for any answers or hints!

N.B. I am aware that this question has already been addressed. In If $f\in L^1_{loc}(\mathbb{R})$ and $\int f\varphi=0$ for all $\varphi$ continuous with compact support, then $f=0$ a.e., I am not quite sure about how to create a sequence of compactly supported continuous functions such that $ \varphi_n\to \frac{f}{|f|+1} $. This particular question may have its answer in If $f\in L^1(\mathbb{R})$ is such that $\int_{\mathbb{R}}f\phi=0$ for all continuous compactly supported $\phi$, then $f\equiv 0$., however here I am unsure about the meaning of a "regularizing sequence"; why does $ \phi_n\ast f\to f $ in $L^1$ sense if $ \phi_n(x) = n\phi(nx) $, where $ \phi\in \mathcal C^\infty_c(\Bbb R) $ with $ \phi\ge 0 $ and $ \int_{\Bbb R}\phi(x)dx=1 $?

Once again, any answer would be much appreciated!

  • $\begingroup$ This statement is true, and known as the fundamental theorem of the calculus of variations, this might help your googling. $\endgroup$
    – Luke
    Sep 4 '17 at 16:11
  • $\begingroup$ In which topology do you consider this limit $\phi_n\to \frac{f}{|f|+1}$ $\endgroup$
    – Guy Fsone
    Sep 4 '17 at 16:12
  • $\begingroup$ @Guy Fsone: I think what is required here is pointwise convergence, so that we may apply the dominated convergence theorem. $\endgroup$
    – Apollo13
    Sep 4 '17 at 16:25
  • $\begingroup$ A good, general tool to know about is a "mollifier". en.wikipedia.org/wiki/Mollifier . This should help with understanding "regularizing sequences". $\endgroup$ Sep 4 '17 at 17:47

First it is well known that, $C^\infty_c$ is dense in $L^p$ for $1\le p<\infty.$

But if $f\in L^p$ so is $\frac{f}{|f|+1}$ because

$$\frac{|f|}{|f|+1} \le |f|$$ therefore there is automatically a sequence $\phi_n$ in $C^\infty_c$ converging $L^p$ to $\frac{f}{|f|+1}$ this solve your first question

Now note that $$ \int_\mathbb R\phi_n(y) dy= \int_\mathbb R \phi(y) dy = 1$$ then, for every $x\in \mathbb R$,

\begin{split} f*\phi_n(x) -f(x) &=&f*\phi_n(x) -f(x)\int_\mathbb R \phi_n(y) dy \\ &= &\int_\mathbb R (f(x-y)-f(x))\phi_n(y) dy\\ &=& n\int_\mathbb R (f(x-y)-f(x))\phi(ny) dy\\ &=& \int_\mathbb R (f(x-\frac{1}{n}y)-f(x))\phi(y) dy \end{split}

Hence assuming $f\in L^1(\mathbb R)$ and using Fubini, we have,

\begin{split} \|f*\phi_n -f\|_1 &=&\int_\mathbb R \int_\mathbb R(f(x-\frac{1}{n}y)-f(x))\phi(y)dxdy\\ &=& \int_\mathbb R \|f(.-\frac{1}{n}y)-f(.)\|_1\phi(y) dy\\ \end{split}

  • $\|f(\cdot-\frac{1}{n}y)-f(\cdot)\|_1|\phi(y)|\le 2\|f(\cdot)\|_1\phi(y) \in L^1(\mathbb R)$ for almost every $y$.
  • And We know the following $\|f(\cdot-\frac{1}{n}y)-f(\cdot)\|_1\to 0$ as $n\to \infty$ for very $f\in L^1(\mathbb R)$ Therefore by convergence dominated theorem we get $$ \|f*\phi_n -f\|_1 \to 0 \qquad , n\to \infty$$

i,e $f*\phi_n \to f$ in $L^1$.

  • 1
    $\begingroup$ Regarding the first part of your answer, I was wondering which $ L^1 $ function would dominate the sequence $ \phi_n \to \frac{f}{|f|+1} $ with $ \phi_n \in C^{\infty}_{c} $, for we want to use the dominated convergence theorem to conclude that $ f = 0 $ a.e. Many thanks! $\endgroup$
    – Apollo13
    Sep 5 '17 at 10:23

For any ball $B(a,r),$ there is a sequence $g_k$ of continuous functions with support in $B(a,r),$ with $|g_k|\le 1$ everywhere, such that $g_k(x) \to \text { sgn }(f(x))$ pointwise a.e. in $B(a,r).$ Thus

$$\tag 1 \int_{B(a,r)} |f| = \int_{B(a,r)} f\cdot \text { sgn }(f) = \lim \int_{B(a,r)} f\cdot g_k =\lim 0 =0.$$

The second equality in $(1)$ follows from the dominated convergence theorem. From $(1)$ we see $f=0$ a.e. on $B(a,r).$ Since $B(a,r)$ was an arbitrary ball, we have $f=0$ a.e. in $\mathbb R^n.$

  • 1
    $\begingroup$ Could you explain why for $ f $ there exists a sequence $ g_k $ with the properties above? Many thanks for your answer. $\endgroup$
    – Apollo13
    Sep 5 '17 at 18:18
  • 1
    $\begingroup$ It follows from Egorov's theorem. It also follows from the density of $C_c(B(a,r))$ in $L^1(B(a,r)).$ Using the latter, choose a sequence $g_k \in C_c(B(a,r))$ such that $g_k \to \text { sgn }(f)$ in $L^1.$ Then some sequence $g_{k_j}$ converges to $\text { sgn }(f)$ a.e. Define $h:\mathbb C\to \mathbb C$ by $h(z)= z, |z|\le 1,$ $h(z) = z/|z|, |z|>1.$ Then $h\circ g_{k_j}$ does the job. $\endgroup$
    – zhw.
    Sep 5 '17 at 19:01
  • $\begingroup$ Thank you very much for the answer! This post helps me a lot. $\endgroup$
    – user682705
    Dec 26 '19 at 19:46
  • $\begingroup$ @user682705 You are welcome, and happy new year! $\endgroup$
    – zhw.
    Dec 26 '19 at 20:06

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