This question has been asked many times. I've read that $\dfrac{d(\delta(f(x)))}{dx}=-f^{'}(x)$. For instance, here, the second answer. I am coming from the context of probability. In particular, I am considering $\lim_{\sigma \rightarrow 0}N(f(x),\sigma)$, and that is how $\delta((f(x)))$ shows up. I would like a reference to a math book or something rigoruous that shows the result presented above, namely that $\dfrac{d(\delta(f(x)))}{dx}=-f^{'}(x)$. I've done some google search, so far I've seen many proofs that are not truly rigoruous, including this; in other posts they say that is so by definition, like here or here, is that so?. That is why I think a book would be ideal, but not necessary nonetheless. Any help would be appreciated.
Asked
Active
Viewed 202 times
0
-
2The posts you refer to are right ... to understand that you first have to know what is the definition of the derivative of a dirac. You should look at a course of distribution theory. – LL 3.14 May 17 '20 at 17:53
-
1Please check distribution (not probability distribution) and derivative of distribution. – user251257 May 17 '20 at 17:53
-
Yes, that's exactly what I am looking for. Any reference in particular? or just any book is fine? – Schach21 May 17 '20 at 17:55
-
2@Schach21 any book on distribution theory will be okay. It is basically the first chapter... or Wikipedia. – user251257 May 17 '20 at 17:56
-
Warning however, $\delta'(f(x)) = - f'(0)$ but what you write here looks like $(\delta(f(x)))' = f(0)' = 0$ – LL 3.14 May 17 '20 at 17:59
-
If it is the limit of a Gaussian law centered in $f(x)$, it should be $\delta_{f(x)}$ in the limit. – LL 3.14 May 17 '20 at 18:09
-
what is difference between $\delta_{f(x)}$ and $\delta(f(x))$, isn't that just notation? let me be more precise. I am considering $\lim_{\sigma \rightarrow 0}N(\mu-f(x),\sigma)$, so I'll get $\delta(\mu-f(x))$. The derivative is with respect to $x$. – Schach21 May 17 '20 at 18:13
-
@Schach21. If you get $\delta(\mu - f(x))$ then you do not get $\delta(f(x)).$ Instead you get $\delta_{f(x)}(\mu).$ – md2perpe May 17 '20 at 18:53
-
Usually the notation $g(f)$ for a distribution $g$ is tbe action of this distribution on a smooth function $f$. In particular, it means $\int gf$ if $g$ is a locally integrable function. – LL 3.14 May 17 '20 at 18:54
-
@md2perpe I made a mistake, I actually do have just $\delta(f(x))$, not with the $\mu$. – Schach21 May 17 '20 at 19:04
-
To me it's still unclear what you really have. You write $N(f(x), \sigma)$ which I interpret as you having $N(f(x), \sigma)(t) = \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{(t-f(x))^2}{2\sigma^2}}.$ – md2perpe May 17 '20 at 19:09
-
Yes, that's exactly what I have. – Schach21 May 17 '20 at 20:11
1 Answers
2
So, now that it has become more clear, let answer more precisely.
Distributions: A distribution $g$ (from the theory of distribution of Laurent Schwartz) is defined as a linear form on the space of compactly supported functions $C^\infty_c$, and we write for every $\varphi∈C^\infty_c$ $$ g(\varphi) = \langle g,\varphi\rangle $$ Locally integrable functions can be seen as distributions by defining $\langle g,\varphi\rangle = ∫ g\varphi$ in this case. Some distributions are not functions, such as the Dirac delta defined by $$ \langle \delta_{a},\varphi\rangle := \varphi(a). $$ The derivative of a distribution $g$ is defined as the distribution $g'$ acting on smooth functions in the following way $$ \langle g',\varphi\rangle := -\langle g,\varphi'\rangle. $$ It is in particular compatible with the case when $g$ is a $C^1$ function by integrating by parts. Moreover, $\langle \delta'_{a},\varphi\rangle := -\varphi'(a).$
You problem: In your case, the normal law centered in $f(x)$, $N(f(x),\sigma)$ converges to the law with distributions $\delta_{f(x)}$. Now we want to compute the derivative with respect to $x$. Let $\varphi\in C^\infty_c$ be a test function. Then (by a theorem similar to the theorem of derivation under the integral sign) $$ \left\langle \frac{\mathrm{d}}{\mathrm{d}x}(\delta_{f(x)}),\varphi\right\rangle = \frac{\mathrm{d}}{\mathrm{d}x}\left\langle \delta_{f(x)},\varphi\right\rangle = \frac{\mathrm{d}}{\mathrm{d}x}\varphi(f(x)) = f'(x)\, \varphi'(f(x)) $$ and by the above definitions $$ \varphi'(f(x)) = \left\langle \delta_{f(x)},\varphi'\right\rangle = -\left\langle \delta_{f(x)}',\varphi\right\rangle. $$ We conclude that $$ \frac{\mathrm{d}}{\mathrm{d}x}(\delta_{f(x)}) = -f'(x)\,\delta_{f(x)}' $$
LL 3.14
- 12,457
-
Thanks! I am a little confused with notation here. At the end, you wrote $\dfrac{d}{dx}(\delta_{f(x)})=-f'(x)\delta'{f(x)}$. Isn't $\dfrac{d}{dx}(\delta{f(x)})=\delta'_{f(x)}$, again this is just notation. – Schach21 May 17 '20 at 21:01
-
Nono, the $x$ is a parameter from the point of view of the dirac. If you want to "think" of the dirac as a function, i.e. $\delta_a(y) = \delta_0(y-a)$ is $0$ everywhere except when $y=a$ (or let say it a very thin Gaussian), then we are computing here $$ \frac{\mathrm{d}}{\mathrm{d}x}(\delta_0(y-f(x))) = - f'(x) \frac{\mathrm{d}}{\mathrm{d}y}(\delta_0(y-f(x))) = - f'(x) \delta'_0(y-f(x)) $$ – LL 3.14 May 17 '20 at 22:20
-
Ok. Got it, but then what is exactly $\delta'_0(y-f(x))$. As you said, $\delta_0(y-a)$ is $0$ erveywhere except when $y=a$, what about $\delta_0'(y-f(x))$? is it $0$ everywhere except at $y=f'(x)$? Sorry for being a little slow. I am reading Functional Analysis by Rudin, the section about distributions and this is starting to make a lot of sense. The idea of a distribution is indeed pretty neat. – Schach21 May 17 '20 at 23:00
-
This is just an image, the dirac can be think of as a function that is zero everywhere except at one point where it is infinity, but since this is not rigorous, the theory of distributions is the good way to formulate that – LL 3.14 May 17 '20 at 23:58
-
If you prefer, you can think about Gaussians very thin, close to a dirac, let say $G_a(y) = G_0(y-a)$ centered in $a$. Then the proof in my above comment works replacing $\delta_0$ by $G_0$ – LL 3.14 May 18 '20 at 00:00
-
Yes, I am aware that I can be thought as function that is zero everywhere except at one point, where it is infnity. What confuses is that you use $\delta$ prime ($\delta'_{f(x)}$), because I don't know what the "prime" there means. – Schach21 May 18 '20 at 00:10
-
This is the derivative (in the sense of distributions for the dirac). In the same way, if $G_0(y) = e^{-y^2}$, then $(G_0(f(x)))' = f'(x)G_0'(f(x))$ where $G_0'(y) = -2y e^{-y^2}$. I defined the derivative of the dirac $\delta_a'$ in my post. – LL 3.14 May 18 '20 at 00:14
-
Of course, $\delta_a'$ is a very singular distribution. It is sometimes used is physics to model a dipole. – LL 3.14 May 18 '20 at 00:19
-
Yes, I read that. I don't exactly know how or what kind of dipole. I only know the dipole formed by two charges of opposite sign. What does that $\delta'_a$ look like? – Schach21 May 18 '20 at 00:24
-
It is $0$ everywhere except in $a$. In $a$, it is like $+\infty$ on the left and $-\infty$ on the right. Once again, it is good to see it as a limit of the derivative of gaussians, or just as $(\delta(x+h)-\delta(x-h))/(2h)$ when $h$ goes to $0$. – LL 3.14 May 18 '20 at 00:27
-
Oh ok, so it is in the traditional sense of derivative. I have one last question, why $\langle \dfrac{d}{dx}\delta_{f(x)},\varphi \rangle =\dfrac{d}{dx}\langle \delta_{f(x)},\varphi \rangle$, in other words, why can you take the derivate outside? because if $\langle g,f \rangle=\int fg dx$, then$ \langle \dfrac{d}{dx} g,f \rangle \neq \dfrac{d}{dx}\langle g,f \rangle$, or am I wrong here?. – Schach21 May 18 '20 at 00:34
-
Yes, once again, $x≠ y$. If you replace $\delta$ by a Gaussian, then $\langle G_{f(x)},\varphi\rangle = ∫G_{f(x)}(y) \varphi(y) ,\mathrm{d} y$. – LL 3.14 May 18 '20 at 03:15
-
-
3 14 In the comments to my question, you said $\delta'(f(x))=f'(0)$, is that $\langle \dfrac{d}{dx}\delta_0, f(x) \rangle$, what does $\delta'f(x)=f'(0)$ mean? The final answer was $f'(x)\delta'{f(x)}$ but that function (?) would be zero everywhere except at $y=f(x)$ from the left where it is $-\infty$ or the right where it is $\infty$, so $f'(x)$ would not matter, that is, it would be the same to have $f'(x)\delta'{f(x)}$ or simply $\delta'_{f(x)}$, which seems strange. The derivative of a distribution should be a distribution, then it is unclear to me what distribution we are getting. – Schach21 May 18 '20 at 14:27
-
No, when I am saying "it look like a function that is $0$ everywhere except ..." this is only a non-rigorous way of having a mental image of the dirac. But $2,\delta_0 ≠ \delta_0$ (one has "integral" $1$ and the other "integral" $2$ for example). In the same way, $2\delta'_a ≠ \delta'_a$. You should really look at a course on distribution theory. The dirac delta is not a function, and this is why distribution theory was created – LL 3.14 May 18 '20 at 19:47
-
Yes, I've been doing that. It'll take me a while to understand. I have one last question. This time I promise it's the last one. It was shown that the derivative was equal to $f'(x)\varphi'(f(x))$. $\varphi$ is a test function, so it only has to be compactly supported and with continuous derivatives. Then, could I just assume that $\varphi$ is the identity in some compact interval, then it vanishes outside, so that $\varphi'(f(x))=1$ in that compact interval? Then the derivative is just $f'(x)$ in that compact interval? That seems highly ilegal, but I don't see why. – Schach21 May 18 '20 at 19:53
-
You can see the $\varphi$ as playing the role of the variable in classical functions. Imagine we are doing your question with a Gaussian ($G_a(y) = G(y-a)$ is a Gaussian centered in $a$). Then we have $\frac{\mathrm{d}}{\mathrm{d}x}G_{f(x)}(y) = - f'(x) G'{f(x)}(y)$. This is a function still depending on $x$ and $y$. In the same way, $\delta{f(x)}(\varphi)$ still depends on the "variables" $x$ and $\varphi$. So of course you can take $y=1$ for your Gaussian, but it will give you only one value of your probability law at a specific point, not the full law. – LL 3.14 May 18 '20 at 20:03
-
In the case of the Gaussian, what you are saying would be like taking $y=0$, and then $\frac{\mathrm{d}}{\mathrm{d}x}G_{f(x)}(0) = - f'(x) G'_{f(x)}(0) = - f'(x) e^{-f(x)^2}$ but I do not think it is what you are looking for – LL 3.14 May 18 '20 at 20:04
-
Ok. I actually think that's what I am looking for. I'd give you even more background motivation (where the whole thing about the limit of the Gaussian came from), but I've bothered you enough. I'll do more reading but I have a better perspective now. Thanks for all the help! – Schach21 May 18 '20 at 20:11