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Let $\gamma_{1}=\mathscr{N}(0,I_{1})$ in $\mathbb{R}$ be the standard Gaussian measure. Consider the sequence $(f_{n})_{n\in\mathbb{N}}\in C_{b}^{1}(\mathbb{R})$ defined by $$f_{n}(x)=\begin{cases} 0,&\mbox{for }x\le 0, \\ nx,&\mbox{for }0\le x \le\frac{1}{n}, \\ 1,&\mbox{for }x\ge\frac{1}{n}. \end{cases}$$

I have already proven that $f_{n}\to f$ in $L^{1}(\mathbb{R},\gamma_{1})$ and that $\sup_{n\in\mathbb{N}}\|f_{n}\|_{W^{1,1}(\mathbb{R},\gamma_{1})}<\infty$, but now I want to prove that $f\notin W^{1,1}(\mathbb{R},\gamma_{1})$.

For every $f\in W^{1,1}(\mathbb{R},\gamma_{1})$, the weak derivative $Df$ must coincide with $\partial f/\partial x$. So I want to prove that this is not the case for our $f$.

We have $f(x)=0$ on the negative half-plane and $f(x)=1$ on the positive half-plane. So we can write $f(x)=1_{\Gamma}(x)$, with $\Gamma$ a positive half line. Then $\partial f/\partial x=0$ since the Dirac function is a generalised function.

$f$ has a weak derivative $Df$ if $$\int_{\mathbb{R}}f\frac{\partial g}{\partial x}d\gamma_{1}=-\int_{\mathbb{R}}g\frac{\partial f}{\partial x}d\gamma_{1}+\int_{\mathbb{R}}f(x)g(x)x\gamma_{1}(dx)$$ holds $\forall g\in C_{b}^{1}(\mathbb{R})$.

Do I now just substitute my $\partial f/\partial x=0$?

In which case, we would have $$\int_{\mathbb{R}}f\frac{\partial g}{\partial x}d\gamma_{1}=\int_{\mathbb{R}}f(x)g(x)x\gamma_{1}(dx)$$

There is a Lemma which states that

For every $f\in C^{1}_{b}(\mathbb{R})$ $$\int_{\mathbb{R}}\frac{\partial f}{\partial x}(x)\gamma_{1}(dx)=\int_{\mathbb{R}}x f(x)\gamma_{1}(dx)$$

Can I apply this somehow to get what I want?

Alex M.
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Jason Born
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  • For a $C^1$ function $g$, is $\nabla g$ the "usual" derivative of $g$? I ask since in Sobolev spaces with respect to Lebesgue measure the derivative is sometimes defined in terms of distributions, i.e. $\nabla g$ satisfies $\int{(\nabla g)h\text{ d}x} = -\int{g\nabla h\text{ d}x}$ for all $h\in C_c^{\infty}$. – Joey Zou Jan 07 '16 at 05:16
  • Also, for ease of reading you may want to say what $f$ actually is. – Joey Zou Jan 07 '16 at 05:38
  • @JoeyZou $f\in C_{b}^{1}(\mathbb{R})$. And we define $\nabla:C_{b}^{1}(\mathbb{R})\to L^{1}(\mathbb{R},\gamma_{1};\mathbb{R})$ by $\partial f/\partial x(x):=\nabla f(x)$. – Jason Born Jan 07 '16 at 13:37
  • Ah, so it's the usual derivative then. And I meant you should say what $f$, the limit of $f_n$ (as you have defined them), is, so that we don't have to guess at a formula for $f$. (That said, I don't think $f$, nor any of the $f_n$'s, are actually $C^1$...) – Joey Zou Jan 07 '16 at 14:30
  • @JoeyZou I've not been able to determine $f$ yet. Or else I would have computed the derivatives by now. – Jason Born Jan 08 '16 at 20:24
  • Then I don't understand your statement "I have already proven $f_n\rightarrow f$ in $L^1(\mathbb{R},\gamma_1)$". Have you shown that there exists some $f$ such that $f_n\rightarrow f$ in $L^1$? (If so, you should really try to figure out what $f$ actually is. It's not too hard to guess at what $f$ should be, if I'm understanding this correctly.) – Joey Zou Jan 09 '16 at 00:09
  • @JoeyZou I used the Dominated Convergence Theorem, since I said that $|f_{n}(x)|\le 1$, although now I see that I may have gone wrong. – Jason Born Jan 09 '16 at 14:51
  • @JoeyZou Is it correct that $\lim_{n\to\infty}f_{n}(x)=1=f(x)$ for $x> 0$ and $f(x)=0$ for $x<0$ and $f_{n}(x)\to\infty$ as $n\to\infty$ for $x=0$? – Jason Born Jan 09 '16 at 14:59
  • Actually, $f_n(0) = 0$ for every $n$, so $f_n(0)\rightarrow 0$, but it doesn't matter much since that's just one point. So $f$ is a step function, which means that its derivative "should be" a Dirac delta, and there are some well known ways to prove that the Dirac delta isn't an $L^1$ function with respect to Lebesgue measure. Not everything will work the same with Gaussian measure since integration by parts works a bit differently, but some of the same ideas should work. Let me know if you are stuck after this point. – Joey Zou Jan 09 '16 at 19:35
  • @JoeyZou Thanks for your help; I get that $f(x)=1_{\Gamma}(x)$, where $\Gamma$ is a positive half line. Then $\partial/\partial x f(x)=0$, since the Dirac function is a generalised function. Surely now it's about proving that $Df\ne 0$? – Jason Born Jan 12 '16 at 20:02

1 Answers1

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Remember that $f \in W^{1,1} (\gamma_1)$ means that $f, f' \in L^1 (\gamma_1)$ where $f'$ is understood in the weak sense. Remember also that $f'$ in the Sobolev sense in the same as $f'$ in the distributional sense, therefore let us compute the latter one. Finally, we shall use that $\gamma_1 = \dfrac 1 {\sqrt {2 \pi}} {\rm e}^{-\frac {x^2} 2}$.

Note, first, that $f = \left\{ \begin{eqnarray} 0 &,& x \le 0 \\ 1 &,& x > 0 \end{eqnarray} \right.$.

Pick $\varphi$ smooth with compact support, arbitrary. We want to compute $f'$ (as a distribution) such that

$$\int f' \varphi {\rm d} \gamma_1 = - \int f \varphi' {\rm d} \gamma_1 + \int f \varphi x {\rm d} \gamma_1 = - \int \limits _0 ^\infty \varphi' {\rm d} \gamma_1 + \int \limits _0 ^\infty \varphi x {\rm d} \gamma_1 = \\ - \int \limits _0 ^\infty \varphi' \frac 1 {\sqrt {2 \pi}} {\rm e}^{-\frac {x^2} 2} {\rm d} x + \int \limits _0 ^\infty \varphi x \frac 1 {\sqrt {2 \pi}} {\rm e}^{-\frac {x^2} 2} {\rm d} x = \\ - \frac 1 {\sqrt {2 \pi}} \left( \left. \varphi {\rm e}^{-\frac {x^2} 2} \right| _0 ^\infty - \int \limits _0 ^\infty \varphi {\rm e}^{-\frac {x^2} 2} (-x) {\rm d} x \right) + \int \limits _0 ^\infty \varphi x \frac 1 {\sqrt {2 \pi}} {\rm e}^{-\frac {x^2} 2} {\rm d} x = \\ \frac {\varphi (0)} {\sqrt {2 \pi}} - \frac 1 {\sqrt {2 \pi}} \int \limits _0 ^\infty \varphi {\rm e}^{-\frac {x^2} 2} x {\rm d} x + \int \limits _0 ^\infty \varphi x \frac 1 {\sqrt {2 \pi}} {\rm e}^{-\frac {x^2} 2} {\rm d} x = \frac {\varphi (0)} {\sqrt {2 \pi}} .$$

Noting that the leftmost-hand term can be explicited as $\frac 1 {\sqrt{2 \pi}} \int f' \varphi {\rm e}^{-\frac {x^2} 2} {\rm d} x$, if you equate the beginning of the previous chain of equalities to its end, you get

$$\int \limits _0 ^\infty f' {\rm e}^{-\frac {x^2} 2} \varphi {\rm d} x = \varphi (0) = \delta (\varphi)$$

and, since this is true for arbitrary $\varphi$, this implies that $f' {\rm e}^{-\frac {x^2} 2} = \delta$ (as distributions, with $\delta$ being Dirac's distribution), whence it follows that $f' = {\rm e}^{\frac {x^2} 2} \delta$.

Remeber now that, if $g$ is some smooth function, then $g \delta = g(0) \delta$, the argument being that

$$\langle g \delta, \varphi \rangle = \langle \delta, g \varphi \rangle = g(0) \varphi (0) = g(0) \delta (\varphi) .$$

Using this, the previous expression for $f'$ can be rewritten as $\color{blue} {f' = \delta}$, which is not a regular distribution (i.e. it does not come from a locally-integrable function), therefore it has no chance of being in $L^1 (\gamma_1)$ (because everything living in $L^1$ should also be locally-integrable).

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Alex M.
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  • Where do you get the formula $\int f'\varphi d\gamma_{1}=-\int f\varphi' d\gamma_{1}$ from, since the integration by parts formula for Gaussian measures that I have is $\int f\varphi' d\gamma_{1}=-\int\varphi f' d\gamma_{1}+\int f\varphi x d\gamma_{1}$? – Jason Born Jan 15 '16 at 17:50
  • @user3482534: Thank you for catching that, it was a serious mistake! It's always relieving to know that one has attentive readers who can catch mistakes. I've fixed it now. – Alex M. Jan 15 '16 at 18:49
  • @Alex M. : your edit to this question http://math.stackexchange.com/revisions/1614692/3 was well intentioned but unfortunately made the notation worse rather than better. (Pinging here because I am not convinced I can ping you on the question I linked). – Carl Mummert Jan 16 '16 at 18:46