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Let $f \in L_{loc}^1 (\mathbb{R}) $ such that $ \int_{\mathbb{R}} f(x) \phi ' (x) dx = 0$, for every $\phi \in C_c ^{\infty}( \mathbb{R}) $. Show that there exist a $c \in \mathbb{R}$ such that $f(x) = c$ a.e.

What I did is:

Assume that $f \in C^ {\infty} $. Therefore, for every $a,b \in \mathbb{R}$, $ 0=\int _a ^b ff' = 1/2 (f(b)^2-f(a)^2)$. Considering that $f$ is continuous, we can conclude that $f(a) = f(b)$.

but i couldn't prove the general case. Any clue?

ianbounos
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2 Answers2

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Suppose that $$ \int_{\mathbb{R}}f(x)\phi'(x)\,\mathrm{d}x=0\tag{1} $$ for all $\phi\in C_c^\infty$. Then consider the functions $$ \psi_{a,\lambda}(x)=\left\{\begin{array}{} 0&\text{for $x\lt a$}\\ \frac{x-a}\lambda&\text{for $x\in[a,a+\lambda]$}\\ 1&\text{for $x\gt a+\lambda$}\tag{2} \end{array}\right. $$ By approximating $\psi_{a,\lambda}-\psi_{b,\mu}$ as close as we want (in $L^\infty$) by $\phi\in C_c^\infty$, we get that $$ \int_{\mathbb{R}}f(x)\phi'(x)\,\mathrm{d}x\tag{3} $$ approximates $$ \frac1\lambda\int_a^{a+\lambda}f(x)\,\mathrm{d}x-\frac1\mu\int_b^{b+\mu}f(x)\,\mathrm{d}x\tag{4} $$ Since $(3)$ is $0$, we get that $(4)$ must be as well. This means that $$ \frac1\lambda\int_a^{a+\lambda}f(x)\,\mathrm{d}x=\frac1\mu\int_b^{b+\mu}f(x)\,\mathrm{d}x\tag{5} $$ Since almost every point is a Lebesgue point of a locally integrable function, $(5)$ means that $$ f(a)=\int_0^1f(x)\,\mathrm{d}x\tag{6} $$ for almost all $a\in\mathbb{R}$.

robjohn
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  • thanks. But I have a problem. I do not see why I can approximate that function in $L_ {\infty}$ because as long as I know $C_c ^{\infty} $ functions are only dense in $L_p$ with $p<\infty$ – ianbounos Jun 29 '17 at 20:12
  • @ianbounos: Note that $\psi_{a,\lambda}-\psi_{b,\mu}\in C_c$. Furthermore, we can approximate by convolution with a scaled and dilated $C_c^\infty$ bump function. – robjohn Jun 29 '17 at 20:24
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clearly $f \in L_{loc}^{1}(\mathbb{R})$ is weak differentiable with weak derivation $Df$since $$-\int_{\mathbb{R}}f\phi' dx =0=\int_{\mathbb{R}}(Df)\phi dx, $$ $ ϕ∈C_{0}^\infty(\mathbb{R}). $ Applying the fundamental theorem of calculus of variation to the right hand side this yields to $Df=0 $ a.e. and therefore $$f \equiv c \in \mathbb{R}$$ a.e.

Simonsays
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  • How do you prove that vanishing weak derivative implies that $f$ is constant a.e.? – Sangchul Lee Jun 29 '17 at 15:22
  • by definition is the weak derivative of $f \in L_1^{loc}(\Omega)$ a function $ \in L_1^{loc}(\Omega)$ – Simonsays Jun 29 '17 at 15:28
  • Weak derivative is initially defined as distribution, it only happens to be realized as measurable function when $f$ has certain regularity strictly stronger than local integrability. (For instance, the Heaviside step function has weak derivative only as distributional sense.) – Sangchul Lee Jun 29 '17 at 15:30
  • concerning your previous question: i used a lemma from my class that stated: if $Du$ is vanishing for $u \in L_1^{loc}$, then u is constant a.e.. Additionally, I think OP is working in the setting of measurable functions right now, but please provide a more general solution, i would be curious. – Simonsays Jun 29 '17 at 15:37
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    What I would like to mention is that $Df$ does not exist as measurable function in general even if $f$ is measurable and locally integrable. Since OP's question does not assume any further regularity on $f$, your answer should be able to cope with this case. As for solution, robjohn's answer shows a nice proof, and I am certain that this is essentially what the lemma taught in your class is telling. – Sangchul Lee Jun 29 '17 at 15:46