# I Distribution Notation

#### joshmccraney

Hi PF!

I am suppose to determine if the following rule is a distribution $$\langle u,\phi \rangle = \int_0^1 \frac{u(x)}{x} \, dx$$ and then also $$\langle u,\phi \rangle = \int_{-\infty}^\infty \phi + 1 \, dx.$$ The notation is throwing me off. At first I thought I had to show $\langle Au + Bu,\phi \rangle = A\langle u,\phi \rangle + B\langle u,\phi \rangle$ but then the next question did not depend on $u$, so I'm a little confused. Can someone clarify the notation please?

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#### Ray Vickson

Homework Helper
Dearly Missed
Hi PF!

I am suppose to determine if the following rule is a distribution $$\langle u,\phi \rangle = \int_0^1 \frac{u(x)}{x} \, dx$$ and then also $$\langle u,\phi \rangle = \int_{-\infty}^\infty \phi + 1 \, dx.$$ The notation is throwing me off. At first I thought I had to show $\langle Au + Bu,\phi \rangle = A\langle u,\phi \rangle + B\langle u,\phi \rangle$ but then the next question did not depend on $u$, so I'm a little confused. Can someone clarify the notation please?
Where is $\phi$ in the right-hand-side of your first example? Where is $u$ in the right-hand-side of your second example? It looks to me like both of your examples involve serious typographical errors.

#### joshmccraney

Where is $\phi$ in the right-hand-side of your first example? Where is $u$ in the right-hand-side of your second example? It looks to me like both of your examples involve serious typographical errors.
This is what I am so confused about. No errors, unless the professor made them when assigning this, which I doubt, but maybe?

http://pi.math.cornell.edu/~web6220/homework/03.html#id2

#### sysprog

From your link: $⟨u, φ⟩=∫_{-∞}^{∞}φ(x)+1\,dx$

From your post: $\langle u,\phi \rangle = \int_{-\infty}^\infty \phi + 1 \, dx.$

Maybe it's a $\TeX$ographical error -- but regardless of which symbol for phi you use, writing "phi" is not the same as writing "phi(x)", right?

Including the stipulation (from your link) that Ω=ℝ, and using your variant of the symbol for lowercase phi, and your TeX code, with its second instance of \phi changed to \phi(x):

$Ω=ℝ$, $\langle u,\phi \rangle = \int_{-\infty}^\infty \phi(x) + 1 \, dx.$

#### stevendaryl

Staff Emeritus
This is what I am so confused about. No errors, unless the professor made them when assigning this, which I doubt, but maybe?

http://pi.math.cornell.edu/~web6220/homework/03.html#id2
It seems like a confusing way to ask the questions. What he might mean is

Let $u$ be that functional that takes another function, $\phi$ and returns $\int_{-\infty}^{+\infty} (\phi(x) + 1) dx$. Is $u$ a distribution?

However, for exercises c and d, the right-hand side involves $u$ and not $\phi$, so I don't see how they can be interpreted as defining $u$. Maybe they're supposed to be defining $\phi$, and asking whether THAT is a distribution?

As I said, a strange way to ask the questions.

#### stevendaryl

Staff Emeritus
I don't know what the meaning of $\mathcal{D}(\Omega)$ is.

#### sysprog

I don't know what the meaning of $\mathcal{D}(\Omega)$ is.
I just googled the $\TeX$ for $\mathcal{D}(\Omega)$, i.e., \mathcal{D}(\Omega).

Apparently $\mathcal{D}(\Omega)$ means a distribution space over an open subset of ${\displaystyle \mathbb {R}^{n}}$.

The following is excerpted from https://en.wikipedia.org/wiki/Distribution_(mathematics)

Basic idea

A typical test function, the bump function Ψ(x).
It is smooth (infinitely differentiable) and has
compact support (is zero outside an interval,
in this case the interval [−1, 1]).

Distributions are a class of linear functionals that map a set of test functions (conventional and well-behaved functions) into the set of real numbers. In the simplest case, the set of test functions considered is D(R), which is the set of functions φ : RR having two properties:
• φ is smooth (infinitely differentiable);
• φ has compact support (is identically zero outside some bounded interval).
A distribution T is a linear mapping T : D(R) → R. Instead of writing T(φ), it is conventional to write ${\displaystyle \langle T,\varphi \rangle }$ for the value of T acting on a test function φ. A simple example of a distribution is the Dirac delta δ, defined by
$${\displaystyle \left\langle \delta ,\varphi \right\rangle =\varphi (0),}$$
meaning that δ evaluates a test function at 0. Its physical interpretation is as the density of a point source.

As described next, there are straightforward mappings from both locally integrable functions and Radon measures to corresponding distributions, but not all distributions can be formed in this manner.

Functions and measures as distributions
Suppose that f : RR is a locally integrable function. Then a corresponding distribution Tf may be defined by
$${\displaystyle \left\langle T_{f},\varphi \right\rangle =\int _{\mathbf {R} }f(x)\varphi (x)\,dx\qquad {\text{for}}\quad \varphi \in D(\mathbf {R} ).}$$
This integral is a real number which depends linearly and continuously on ${\displaystyle \varphi }$. Conversely, the values of the distribution Tf on test functions in D(R) determine the pointwise almost everywhere values of the function f on R. In a conventional abuse of notation, f is often used to represent both the original function f and the corresponding distribution Tf. This example suggests the definition of a distribution as a linear and, in an appropriate sense, continuous functional on the space of test functions D(R).

Similarly, if μ is a Radon measure on R, then a corresponding distribution Rμ may be defined by
$${\displaystyle \left\langle R_{\mu },\varphi \right\rangle =\int _{\mathbf {R} }\varphi \,d\mu \qquad {\text{for}}\quad \varphi \in D(\mathbf {R} ).}$$
This integral also depends linearly and continuously on ${\displaystyle \varphi }$, so that Rμ is a distribution. If μ is absolutely continuous with respect to Lebesgue measure with density f and dμ = f dx, then this definition for Rμ is the same as the previous one for Tf, but if μ is not absolutely continuous, then Rμ is a distribution that is not associated with a function. For example, if P is the point-mass measure on R that assigns measure one to the singleton set {0} and measure zero to sets that do not contain zero, then
$${\displaystyle \int _{\mathbf {R} }\varphi \,dP=\varphi (0),}$$
so that RP = δ is the Dirac delta.

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#### joshmccraney

I just googled the $\TeX$ for $\mathcal{D}(\Omega)$, i.e., \mathcal{D}(\Omega).

Apparently $\mathcal{D}(\Omega)$ means a distribution space over an open subset of ${\displaystyle \mathbb {R}^{n}}$.

The following is excerpted from https://en.wikipedia.org/wiki/Distribution_(mathematics)

Basic idea

View attachment 239168

A typical test function, the bump function Ψ(x).
It is smooth (infinitely differentiable) and has
compact support (is zero outside an interval,
in this case the interval [−1, 1]).

Distributions are a class of linear functionals that map a set of test functions (conventional and well-behaved functions) into the set of real numbers. In the simplest case, the set of test functions considered is D(R), which is the set of functions φ : RR having two properties:
• φ is smooth (infinitely differentiable);
• φ has compact support (is identically zero outside some bounded interval).
A distribution T is a linear mapping T : D(R) → R. Instead of writing T(φ), it is conventional to write ${\displaystyle \langle T,\varphi \rangle }$ for the value of T acting on a test function φ. A simple example of a distribution is the Dirac delta δ, defined by
$${\displaystyle \left\langle \delta ,\varphi \right\rangle =\varphi (0),}$$
meaning that δ evaluates a test function at 0. Its physical interpretation is as the density of a point source.

As described next, there are straightforward mappings from both locally integrable functions and Radon measures to corresponding distributions, but not all distributions can be formed in this manner.

Functions and measures as distributions
Suppose that f : RR is a locally integrable function. Then a corresponding distribution Tf may be defined by
$${\displaystyle \left\langle T_{f},\varphi \right\rangle =\int _{\mathbf {R} }f(x)\varphi (x)\,dx\qquad {\text{for}}\quad \varphi \in D(\mathbf {R} ).}$$
This integral is a real number which depends linearly and continuously on ${\displaystyle \varphi }$. Conversely, the values of the distribution Tf on test functions in D(R) determine the pointwise almost everywhere values of the function f on R. In a conventional abuse of notation, f is often used to represent both the original function f and the corresponding distribution Tf. This example suggests the definition of a distribution as a linear and, in an appropriate sense, continuous functional on the space of test functions D(R).

Similarly, if μ is a Radon measure on R, then a corresponding distribution Rμ may be defined by
$${\displaystyle \left\langle R_{\mu },\varphi \right\rangle =\int _{\mathbf {R} }\varphi \,d\mu \qquad {\text{for}}\quad \varphi \in D(\mathbf {R} ).}$$
This integral also depends linearly and continuously on ${\displaystyle \varphi }$, so that Rμ is a distribution. If μ is absolutely continuous with respect to Lebesgue measure with density f and dμ = f dx, then this definition for Rμ is the same as the previous one for Tf, but if μ is not absolutely continuous, then Rμ is a distribution that is not associated with a function. For example, if P is the point-mass measure on R that assigns measure one to the singleton set {0} and measure zero to sets that do not contain zero, then
$${\displaystyle \int _{\mathbf {R} }\varphi \,dP=\varphi (0),}$$
so that RP = δ is the Dirac delta.
Yea I read this too prior to posting. It does not address the professor's HW statement, which is making me think he has a typo. I will keep everyone posted on his response (I emailed him).

Also, it's unclear to me what $D'(\mathbb R)$ means; anyone have an idea?

#### sysprog

joshmccraney 428835 said:
Also, it's unclear to me what $D'(\mathbb R)$ means; anyone have an idea?
By the definitions in the Symbols list immediately after the Contents at the start of:

NASA Technical Paper 3428
Introduction to Generalized Functions With Applications in Aerodynamics and Aeroacoustics
F. Farassat, Langley Research Center $\bullet$ Hampton, Virginia, Corrected Copy (April 1996)

$\mathcal{D}$ -- space of infinitely differentiable functions with bounded support (test functions)
$\mathcal{D'}$ -- space of generalized functions based on $\mathcal{D}$
$\Omega$ -- open interval or region of space; $\mathcal{\delta}\Omega$ -- boundary of $\Omega$

From the text:
The space of generalized functions on $\mathcal{D}$ is denoted $\mathcal{D'}$ . Figure 1 shows schematically how we extended the space of ordinary functions to generalized functions. We call ordinary functions regular generalized functions, whereas all other generalized functions (such as the Dirac delta function) are called singular generalized functions.
$$F[\phi]=\int^{+\infty}_{-\infty} f(x)\phi(x)\ dx$$
I found that paper to be well worth reading. I really appreciate, especially in a mathematical paper the title of which begins with "Introduction to", the up-front list of symbols. Even if you know the meanings of most of the symbols in such a paper, if you encounter something unfamiliar, you can find yourself struggling to hold onto the concepts while you look up the unknown symbol. With such a good symbol list so easily found that's of much less concern. The paper is very clear, and without condescension, is insightfully explanatory.

From my interpretation of other sources:

${\displaystyle \mathbb {R}^{n}}$ -- a real co-ordinate space over $n$ dimensions
$\mathcal{D}(\Omega)$ -- a distribution space over an open subset of ${\displaystyle \mathbb {R}^{n}}$
$\mathcal{D'}(\Omega)$ -- the space (not necessarily strongly) dual to $\mathcal{D}(\Omega)$
$\mathcal{D}=\mathcal{D}({\displaystyle \mathbb {R}^{n}})$ -- stipulation or default for $\mathcal{D}$
$\mathcal{D'}=\mathcal{D'}({\displaystyle \mathbb {R}^{n}})$ -- stipulation or default for $\mathcal{D}$
$\Omega={\displaystyle \mathbb {R}}$ -- $\Omega$ is the subset ${\displaystyle \mathbb {R}}$ of ${\displaystyle \mathbb {R}^{n}}$ such that $n=1$
$\Omega=(0, 1)$ -- $\Omega$ is the (unit interval) $[1, 0]$ subset of ${\displaystyle \mathbb {R}}$

And based on the foregoing:
$\mathcal{D'}(\mathbb R)$ -- the space of generalized functions acting distributionally over ${\displaystyle \mathbb {R}}$

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#### stevendaryl

Staff Emeritus
Thanks for clearing that up. But the question is: what does the instructor mean when he says something like:

$\langle u, \phi \rangle =$ something involving $u$ but not $\phi$, or something involving $\phi$ but not $u$?

If the right side of the $=$ only involves $\phi$, it might be interpreted as implicitly defining $u$ through showing it's action on $\phi$. But then if the right side only involves $u$, is that implicitly defining $\phi$?

#### sysprog

Thanks for clearing that up. But the question is: what does the instructor mean when he says something like:

$\langle u, \phi \rangle =$ something involving $u$ but not $\phi$, or something involving $\phi$ but not $u$?

If the right side of the $=$ only involves $\phi$, it might be interpreted as implicitly defining $u$ through showing it's action on $\phi$. But then if the right side only involves $u$, is that implicitly defining $\phi$?
Looks like the suspicions were well-founded:

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#### stevendaryl

Staff Emeritus
Looks like the suspicions were well-founded:
Okay. Then in that case, we're implicitly defining $u$.

#### sysprog

Okay. Then in that case, we're implicitly defining $u$.
Right -- and regarding your remaining question, Ray Vickson and the OP were right about there being a typo.

#### joshmccraney

Thanks all! Yep, with the typo corrected, it all makes sense. However, I am wondering if I am doing something wrong. Specifically, looking at c) I was thinking $$\langle u, A\phi + B\phi \rangle = \int_0^2 \frac{A\phi + B\phi}{1-x}\, dx = A\langle u, \phi \rangle+B\langle u, \phi \rangle$$
so it is linear, but since there is a singularity at $x=1 \in \Omega$ the distribution is not generally continuous in all $\Omega$, yet since for part d) the distribution is continuous everywhere in $\Omega = (0,1)$ (since the sinularity is at $x=1 \notin \Omega$) and hence is a distribution.

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#### sysprog

You'll presumably recall that in general, an open interval is conventionally designated as open by use of parentheses, as in $\Omega = (0, 1)$ as distinguished from brackets, as in ${\displaystyle \mathbb \Omega\ =[0,1]}$, and that an open interval does not include its boundaries, whereas a closed interval does include its boundaries.

In the case in which $\Omega =(0, 1)$ means that $\Omega$ is the non-endpoint-inclusive unit interval proper subset of ${\displaystyle \mathbb {R}}$ (where ${\displaystyle \mathbb {R}^{n}}={\displaystyle \mathbb {R}^1}={\displaystyle \mathbb {R}}$), the boundaries are the endpoints, 0 and 1.

I would interpret the open subset $\Omega=(0, 2)$ of ${\displaystyle \mathbb {R}}$ interval as inclusive of 1 (1 not being an endpoint in that interval), but if you map it as the pair of intervals, (0, 1) and (1, 2), you've (in my view) excluded 0, 1, and 2.

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#### joshmccraney

You'll presumably recall that in general, an open interval is conventionally designated as open by use of parentheses, as in $\Omega = (0, 1)$ as distinguished from brackets, as in ${\displaystyle \mathbb \Omega\ =[0,1]}$, and that an open interval does not include its boundaries, whereas a closed interval does include its boundaries.

In the case in which $\Omega =(1, 0)$ means that $\Omega$ is the non-endpoint-inclusive unit interval proper subset of ${\displaystyle \mathbb {R}}$ (where ${\displaystyle \mathbb {R}^{n}}={\displaystyle \mathbb {R}^1}={\displaystyle \mathbb {R}}$), the boundaries are the endpoints, 0 and 1.

I would interpret the open subset $\Omega=(0, 2)$ of ${\displaystyle \mathbb {R}}$ interval as inclusive of 1, 1 not being an endpoint in that interval, but if you map it as the pair of intervals, (0, 1) and (1, 2), you've (in my view) excluded 0, 1, and 2.
It sounds like you're agreeing with me then?

#### sysprog

I would not interpret $\Omega = (0, 2)$ as $\Omega = (0, 1) + (1, 2)$, as that would in my view exclude 1, where $(0, 2)$ would not in my view exclude 1.

#### joshmccraney

I would not interpret $\Omega = (0, 2)$ as $\Omega = (0, 1) + (1, 2)$, as that would in my view exclude 1, where $(0, 2)$ would not in my view exclude 1.
Right, me neither. Isn't that why c) is not a distribution yet d) is? Thought that's what I was saying.

#### sysprog

I agree with your conclusion, but please don't take my word for it; please instead refer to your own (in my view reasonable) interpretation of what "continuous" means in the definition of "distribution" in the footnote in the problem statement.

#### joshmccraney

I agree with your conclusion, but please don't take my word for it; please instead refer to your own (in my view reasonable) interpretation of what "continuous" means in the definition of "distribution" in the footnote in the problem statement.
Thanks. I'll see if anyone else posts. I think this is the correct approach though.

#### WWGD

Gold Member
Thanks all! Yep, with the typo corrected, it all makes sense. However, I am wondering if I am doing something wrong. Specifically, looking at c) I was thinking $$\langle u, A\phi + B\phi \rangle = \int_0^2 \frac{A\phi + B\phi}{1-x}\, dx = A\langle u, \phi \rangle+B\langle u, \phi \rangle$$
so it is linear, but since there is a singularity at $x=1 \in \Omega$ the distribution is not generally continuous in all $\Omega$, yet since for part d) the distribution is continuous everywhere in $\Omega = (0,1)$ (since the sinularity is at $x=1 \notin \Omega$) and hence is a distribution.
Remember that the definition of continuity for Distributions is that of Sequential continuity, i.e., the distribution f is continuous off for all test functions $\phi$ ( $C^{ \infty}$ of compact support)_ is continuous iff(Def.) $( \phi_n \rightarrow \phi ) \rightarrow ( f(\phi_n) \rightarrow f(\phi))$ Remember that continuity always implies sequential continuity, but not viceversa ( unless your space is topolologicall 1st countable). You also need to know the specific definition of convergence for/in the space of Test Functions: these must each have compact support in the same set and have the n-th derivative $\phi_k^{(n)} \rightarrow \phi^{(n)}$, where (n) means n-th derivative. And the convergence must be uniform. A lot of machinery, but when you finally get it, it simplifies things a lot.

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