A Heisenberg equations of Klein-Gordon Field in Space-Time

[JD_PM]

4) And the most important thing is the following: Since \varphi \mapsto (-1)^{|m|} \partial^{m}\varphi is linear and continuous (from \mathcal{D}(\mathbb{R}^{n}) \ \mbox{into} \ \mathcal{D}(\mathbb{R}^{n})), then (1) can be used to show that the operation of differentiation f \mapsto \partial^{m}f is linear and continuous (from \mathcal{D}^{\prime}(\mathbb{R}^{n}) \ \mbox{into} \ \mathcal{D}^{\prime}(\mathbb{R}^{n})).

I think I know how to show linearity; given the differential operator $D^{\prime}:\mathcal{D}^{\prime}(\mathbb{R}^{n}) \rightarrow \mathcal{D}^{\prime}(\mathbb{R}^{n}):f \mapsto \partial^{m}f$, we let $\varphi_1, \varphi_2 \in \mathcal{D}^{\prime}(\mathbb{R}^{n})$, $\alpha, \beta \in \mathbb{C}$ and get

$$D^{\prime}f =\partial^m f$$

Multiplying the above equation both sides by $(\alpha \varphi_1+ \beta \varphi_2)$ (on the right) allows us to use the well-known definition to show linearity

$$D^{\prime}f (\alpha \varphi_1+ \beta \varphi_2) =\partial^m f (\alpha \varphi_1+ \beta \varphi_2)=(-1)^m f\partial^m (\alpha \varphi_1+ \beta \varphi_2)$$ $$=\alpha (-1)^m f\partial^m ( \varphi_1)+\beta (-1)^m f\partial^m ( \varphi_1)=\alpha f(\varphi_1)+\beta f(\varphi_2)$$

Where I've used the fact that $D:\mathcal{D}(\mathbb{R}^{n}) \rightarrow \mathcal{D}(\mathbb{R}^{n}):\varphi \mapsto (-1)^{|m|}\partial^{m} \varphi$ is linear.

Argh I'm having trouble to show continuity though.

 

[martinbn]

Argh I'm having trouble to show continuity though.

Are you studying some particular book/notes or just post in this thread? I may have missed it, but what continuous means for distributions wasn't said in this thread.

 

[JD_PM]

Hi martinbn

Are you studying some particular book/notes or just post in this thread? I may have missed it, but what continuous means for distributions wasn't said in this thread.

I was trying to show what samalkhaiat wrote here

4) And the most important thing is the following: Since \varphi \mapsto (-1)^{|m|} \partial^{m}\varphi is linear and continuous (from \mathcal{D}(\mathbb{R}^{n}) \ \mbox{into} \ \mathcal{D}(\mathbb{R}^{n})), then (1) can be used to show that the operation of differentiation f \mapsto \partial^{m}f is linear and continuous (from \mathcal{D}^{\prime}(\mathbb{R}^{n}) \ \mbox{into} \ \mathcal{D}^{\prime}(\mathbb{R}^{n})).

I showed linearity but got stuck in showing continuity.

 

[martinbn]

Hi martinbn
I was trying to show what samalkhaiat wrote here
I showed linearity but got stuck in showing continuity.

What does it mean to be continuous?

 

[JD_PM]

What does it mean to be continuous?

3) A distribution (or generalized function) f is a continuous functional on \mathcal{D}(\mathbb{R}^{n}). That is, if \varphi_{k} \to \varphi as k \to \infty in \mathcal{D}(\mathbb{R}^{n}), then (f , \varphi_{k}) \to (f , \varphi ) as k \to \infty.

 

[martinbn]

3) A distribution (or generalized function) f is a continuous functional on \mathcal{D}(\mathbb{R}^{n}). That is, if \varphi_{k} \to \varphi as k \to \infty in \mathcal{D}(\mathbb{R}^{n}), then (f , \varphi_{k}) \to (f , \varphi ) as k \to \infty.

In short, distribution is a continuous linear functional on the spaces of “nice” functions.

And what does \varphi_{k} \to \varphi as k \to \infty mean?

 

[JD_PM]

Honestly I still do not completely get it either

 

[martinbn]

Honestly I still do not completely get it either

My point was, if you have not even seen the definition how do you expect to understand or prove the continuity!?

 

[samalkhaiat]

$\mathcal{D}(\mathbb{R}^{n})$ is the vector space of operators $\varphi: \Bbb R^n \rightarrow \Bbb R^n$ having the following two properties: 1) $\varphi$ is smooth and 2) $\varphi$ has compact support.

\mathcal{D}(\mathbb{R}^{n}) is the space of all infinitely differentiable functions \varphi : \mathbb{R}^{n} \to \mathbb{R} with compact support. The support \mbox{supp} \ \varphi (x) of a function \varphi(x) is the closure of the set on which \varphi (x) \neq 0.

Mmm I've got a little doubt here: you said $(f,\varphi)$ is a complex number, but as we're dealing with $L_f:\mathcal{D}(\mathbb{R}^{n}) \rightarrow \Bbb R$ it would make more sense to me that $(f,\varphi) \in \Bbb R$ but of course I may be missing something really evident here).

No, that is a classic misunderstanding. When we turn the set \mathcal{D}(\mathbb{R}^{n}) into a vector space over the field of complex numbers, the elements of \mathcal{D}(\mathbb{R}^{n}) become complex-valued functions of the real variables x = (x^{1}, \cdots , x^{n}) in the open set \mathcal{O} \subset \mathbb{R}^{n}. Let \alpha and \beta be two complex numbers and write \varphi = \alpha \varphi_{1} + \beta \varphi_{2}. Now, is the number (f , \varphi ) real or complex?

For a locally integrable function f(x), the functional F_{f} [\varphi] \equiv (f , \varphi) = \int d^{n}x \ f(x) \varphi (x) is linear in that, if \varphi_{1} and \varphi_{2} are in \mathcal{D} and if \alpha and \beta are two (real or complex) constants, then F_{f}[\alpha \varphi_{1} + \beta \varphi_{2}] = \alpha \ F_{f}[\varphi_{1}] + \beta \ F_{f}[\varphi_{2}].

 

[samalkhaiat]

OK the idea I have in mind is showing that we get the same result either using $\partial^{k} \left( \partial^{m}f \right)$ or $\partial^{m} \left( \partial^{k} f \right)$

When $\partial^{k} \left( \partial^{m}f \right)$ we get

$$\int dx \ \partial^{k} \left( \partial^{m}f \right) \ \varphi (x) = \partial^{k} \left( \partial^{m-1}f \right) \ \varphi (x) -\int dx \partial^{k} \left( \partial^{m-1}f \right) \ \partial \varphi (x)$$

Applying integration by parts $m$ times one gets

I am afraid that was garbage. For \varphi \in \mathcal{D}, use the following equalities \partial^{r + s} \varphi = \partial^{r}(\partial^{s}\varphi) = \partial^{s}(\partial^{r}\varphi) , integrate with f \in \mathcal{D}^{\prime} and multiply the equalities with (-1)^{r + s}: (-1)^{r + s} (f , \partial^{r + s} \varphi ) = (-1)^{r + s}(f ,\partial^{r}(\partial^{s}\varphi) ) = (-1)^{r + s}(f , \partial^{s}(\partial^{r}\varphi) . Now, use our definition (1) once on the left-hand-side and twice in the middle and the right-hand sides. This gives you (\partial^{r + s}f , \varphi ) = ( \partial^{s}(\partial^{r}f) , \varphi ) = (\partial^{r}(\partial^{s}f) , \varphi ). QED.

 

[samalkhaiat]

Honestly I still do not completely get it either

The functional F_{f} [\varphi] is continuous in the following sense: Consider a sequence of functions \{ \varphi_{k}(x) \} in \mathcal{D} (\mathbb{R}^{n}) with the following two properties:
1. There exists a compact set K \subset \mathbb{R}^{n} such that for all k, \mbox{supp}\ \varphi_{k} \subset K.
2. \lim_{k \to \infty} \partial^{m}\varphi_{k}(x) = 0 uniformly for all m = 0, 1, 2, \cdots.
Such a sequence is said to go to 0 in \mathcal{D} and is written \varphi_{k} \to 0, \ k \to \infty in \mathcal{D}.
We then say that the functional F_{f}[\varphi] is continuous if F_{f}[\varphi_{k}] \to 0 for \varphi_{k} \to 0, \ k \to \infty in \mathcal{D}.

Let us discuss this definition of continuity of linear functionals on the space \mathcal{D}. Continuity is a topological property. The space \mathcal{D} is a linear vector space. It is made into a topological vector space by defining the neighbourhood of the “point” \varphi (x) = 0 by a sequence of semi-norms. One can show (and that one is definitely not you) that the two conditions required above in the definition of \varphi_{k} \to 0 in \mathcal{D} follow from the conditions used to define the neighbourhood of \varphi(x) = 0. The definition of continuity of linear functional on \mathcal{D} can then be based on the weak or strong topologies of space \mathcal{D}^{\prime}. It so happens that the definitions of continuity based on these topologies are equivalent to the above definition of F_{f}[\varphi_{k}] \to 0 if and only if \varphi_{k} \to 0 in \mathcal{D}. Furthermore, since \mathcal{D} is a linear space, we can define \varphi_{k} \to \varphi , \ k \to \infty in \mathcal{D} if (\varphi_{k} - \varphi ) \to 0, \ k \to \infty in \mathcal{D}, and because F_{f}[\varphi] is linear, we can also say that F_{f} [\varphi] is continuous on \mathcal{D} if when \varphi_{k} \to \varphi, we have F_{f} [\varphi_{k}] \to F_{f} [\varphi].

Now, if you have digest the above, proving that f \mapsto \partial^{m}f is continuous from \mathcal{D}^{\prime} into \mathcal{D}^{\prime} is “easy”. First, recall our definition (\partial^{b}f , \varphi ) = (-1)^{|m|} (f , \partial^{m}\varphi ) . \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1) Since \varphi \mapsto (-1)^{|m|}\partial^{m}\varphi is linear and continuous, the functional \partial^{m}f defined by the right-hand-side of (1) is a generalized function in \mathcal{D}^{\prime}(\mathbb{R}^{n}), i.e., the derivative of a distribution is also a distribution. Now, suppose f_{k} \to 0, \ k \to \infty in \mathcal{D}^{\prime}(\mathbb{R}^{n}). Then for all \varphi \in \mathcal{D}(\mathbb{R}^{n}) we have (\partial^{m}f_{k} , \varphi ) = (-1)^{|m|} (f_{k} , \partial^{m}\varphi ) \to 0, \ k \to \infty , meaning that \partial^{m}f_{k} \to 0, \ k \to \infty in \mathcal{D}^{\prime}(\mathbb{R}^{n}). QED

 

[Alfred1999]

3) Given a generalized function f \in \mathcal{D}^{\prime}(\mathbb{R}^{n}) and a good function a \in C^{\infty}(\mathbb{R}^{n}), how do you know that the Leibniz rule holds for the differentiation of the product af? You don’t. But, if you use the definition (1), you can show that \partial^{m}\left( af \right) = \sum_{k \leq m} \begin{pmatrix} k \\ m \end{pmatrix} \ \partial^{k}a \ \partial^{m - k}f .

To properly answer this, first let's set up the proper definitions so everything checks out latter.
Let V be a vector space over the scalar field F which can be either \mathbb{R} or \mathbb{C}. Given two elements of V such as u, v \in V be elements of V, then we define the inner product as map

$$(.,.): V \times V \to F$$
$$ u, v \to (u, v) \in F$$

such that it satisfies the following properties:

1. Conjugate symmetry:​

$$(u, v) = (u, v)^{*}$$

where the notation means that given z \in \mathbb{C}; then z^{*} is just the complex-conjugate of z.​

2. Linearity in the first argument:​

$$\alpha, \beta \in F; \quad u, v, b \in V; \quad (\alpha u + \beta v, b) = \alpha(u, b) + \beta(v, b)$$

3. Positive definite:​

$$(x, x) \ge 0 \quad \forall x \in V \setminus \{0\}$$

Now, let \mathcal{D}(\mathbb{R}^n) be the vector space over the field \mathbb{R} (if it was over the field \mathbb{C}, then we'd say \mathcal{D}(\mathbb{C}^n)) such that its elements are "nice" functions \varphi :\mathbb{R}^n \to \mathbb{R}. We say that a function is "nice" if:

1. It's smooth
2. It has compact support

And finally, given f, g \in \mathcal{D}(\mathbb{R}^n), we take the our inner product to be:
$$(f, g) := \displaystyle\int d^n x \bigg( f(x) g(x) \bigg)$$

With all this in place, we can finally define a "distribution". If V = \mathcal{D}(\mathbb{R}^n) and V^{*} is its dual space, then T_f \in V^{*} is a linear form such that if f, \phi \in V are nice functions, then \quad T_f[\phi] = (f, \phi) \in \mathbb{R}. We call T_f a "distribution", but usually we abuse the notation by calling f itself the distribution. Because of that, it is desirable to define the derivative of a distribution such that \partial (T_f) = T_{\partial f}. Using integration by parts and the fact that the nice functions have compact support:

$$(\partial f, \phi) = \displaystyle\int d^n x \bigg( \partial f(x) \phi(x) \bigg) = \cancel{\bigg[ f(x) \phi(x) \bigg]_{-\infty}^{\infty}} - \displaystyle\int d^n x \bigg( f(x) \partial \phi(x) \bigg) = -(f, \partial \phi)$$

$$\implies T_{\partial f}[\phi] = -T_{f}[\partial \phi]$$

From now on we'll abuse the notation too and refer to the distribution T_f just as f.

Now, before proving the statement, I would like to point out that there's an error in the binomial number that you've written there, the m and k are swapped. I think the correct formula is:

$$\partial^{m}\left( af \right) = \sum_{k \leq m} \begin{pmatrix} m \\ k \end{pmatrix} \ \partial^{k}a \ \partial^{m - k}f = \sum_{k = 0}^m \begin{pmatrix} m \\ k \end{pmatrix} \ \partial^{k}a \ \partial^{m - k}f$$

based on what I see on wikipedia here.

Anyway, let's begin with the proof:
We'll first prove that the distribution f satisfies the Leibniz rule as usual \partial(af) = (\partial a) f + a(\partial f) . Then the proof for the given formula for a function a and a functional f is the same as it would be for two functions.

Let f(x), \varphi(x) \in V be nice functions and f be a distribution. Also let \varphi(x) = a(x) \phi(x) where a, \phi \in V. We'll understand that when we write f(x) we refer to the function and when we write f or f[.] we refer to the distribution.

$$f[\varphi] = (f(x), \phi(x)) = \displaystyle\int d^n x \bigg( f(x) \varphi(x) \bigg) = \displaystyle\int d^n x \bigg( a(x) f(x) \phi(x) \bigg) = $$

$$ = \displaystyle\int d^n x \bigg( [a(x) f(x)] \phi(x) \bigg) = (af)[\phi] \equiv g[\phi]$$

where we have defined the distribution g as af and the function g(x) := a(x)f(x). Now

$$(\partial f, \varphi) = \displaystyle\int d^n x \bigg( \partial f(x) \varphi(x) \bigg) = -\displaystyle\int d^n x \bigg( f(x) \partial\{ a(x) \phi(x) \} \bigg) = $$

$$= -\displaystyle\int d^n x \bigg( \partial a(x) f(x) \phi(x) \bigg) -\displaystyle\int d^n x \bigg( g(x) \partial \phi(x) \bigg) = -(\partial a(x) f(x), \phi(x)) -(g(x), \partial \phi(x))$$

$$(\partial f, \varphi) = \displaystyle\int d^n x \bigg( \partial f(x) \varphi(x) \bigg) = \displaystyle\int d^n x \bigg( [a(x) \partial f(x)] \phi(x) \bigg) = (a(x)\partial f(x), \phi(x))$$

$$(a\partial f)[\phi] = -(\partial af)[\phi] - g[\partial \phi] = -(\partial af)[\phi] + (\partial g)[\phi] \implies$$

$$\implies \bigg( \partial(af) \bigg) = \bigg( (\partial a) f \bigg) + \bigg(a(\partial f) \bigg)$$

and so we've proven that distributions obey the Leibniz rule.

Q.E.D.​

And so, because distributions behave the same way under differentiation as functions do, you can copy-paste the proof for the formula

$$\partial^{m}\left( af \right) = \sum_{k \leq m} \begin{pmatrix} m \\ k \end{pmatrix} \ \partial^{k}a \ \partial^{m - k}f = \sum_{k = 0}^m \begin{pmatrix} m \\ k \end{pmatrix} \ \partial^{k}a \ \partial^{m - k}f$$

for a and f being functions (which is trivial) into the one where they are distributions; thus proving the formula for distributions as well.
You can check that proof here.