If we sum the Dirac delta function with a constant, what is the result? I.e., $k+\delta(x)$, where $k$ is a constant.
Asked
Active
Viewed 500 times
0
-
Think about the values that the Delta function takes on over the real line and at $x=0$. – aleden Jan 07 '19 at 13:30
-
something like: $+\infty$ when $x=0$ and $0$ in everywhere else? – ARF Jan 07 '19 at 13:40
-
3The Dirac delta isn't a function. That being said, I don't see why anything special at all would happen. $\delta(x) + k$ is just $\delta(x) + k$, and that's it. – Arthur Jan 07 '19 at 13:42
-
I would like to plot the result and I would like to know whether $k+\delta(x)\approx k$ is correct – ARF Jan 07 '19 at 13:52
-
I agree with @Arthur , $\delta +k$ is the functional on the space of test functions $\varphi$ such that $\langle \delta +k,\varphi\rangle=\varphi(0)+k$ or the measure such that $\int \varphi d(\delta +k)=\int\varphi d\delta +k$ – Peter Melech Jan 07 '19 at 13:55
-
1@PeterMelech No, $<\delta +k,\phi>$ should be $\phi(0)=k\int\phi$. (It certainly can't be $\phi(0)+k$; that's not linear in $\phi$.) – David C. Ullrich Jan 07 '19 at 14:23
-
@DavidUllrich This means that the right answer is $\delta (x)+k=k$ since $\phi=1$ – ARF Jan 07 '19 at 14:34
-
@David Sure that wouldn't be a functional, You're right. And then what is $k\delta$ ? – Peter Melech Jan 07 '19 at 14:48
-
@ARF Huh??? That doesn't make any sense that II can see – David C. Ullrich Jan 07 '19 at 14:53
-
1@PeterMelech The pairing is linear : $<k\delta,\phi>=k\phi(0)$. – David C. Ullrich Jan 07 '19 at 14:54
-
@David I see, that makes sense – Peter Melech Jan 07 '19 at 14:58
-
@David How possible $k \int_{-\infty}^{+\infty}\phi=\phi(0)$? – ARF Jan 07 '19 at 16:19
-
@ARF Why do you imagine that $k\int\phi=\phi(0)$ follows from anything I said? – David C. Ullrich Jan 07 '19 at 16:23
1 Answers
3
There's a lot of confusion here. The "delta function" is sometimes treated "informally", as though it were a function with the property that $$\int f(x)\delta(x)=f(0)$$for all $f$. There may be some merit to such an "informal" presentation in some contexts, but in fact it's wrong; there is no such function.
$\newcommand\D{\mathcal D}$
If we want to get this straight we alas need to start with a more formally correct definition. First, $\D$ is the space of all infinitely differentiable functions with compact support. A distribution is a map $u:\D\to\Bbb R$ such that (i) $u$ is linear (ii) $u$ is continuous.
Explaining exactly what it means to say $u$ is continuous gets a little complicated. We don't need to get that straight here, we just do need to keep in mind that a distribution is not a function (ie not a function defined on the line), rather a distribution is a certain sort of linear mapping from $\D$ to $\Bbb R$.
The space of distributions is denoted $\D'$. So if $u\in\D'$ and $\phi\in\D$ then $$u(\phi)\in\Bbb R.$$The notation $$<u,\phi>:=u(\phi)$$is often convenient.
If $f$ is a (locally integrable) function on $\Bbb R$ then $f$ may be regarded as a distribution; here the distribution corresponding to the function $f$ is defined by $$<f,\phi>=\int f\phi.$$(Where $\int$ means $\int_{-\infty}^\infty$.)
If $u$and $v$ are distributions and $k$ is a constant then the distributions $u+v$ and $ku$ are defined by $$<u+v,\phi>=<u,\phi>+<v,\phi>$$and $$<ku,\phi>=k<u,\phi>.$$
And the distibution $\delta$ is defined by $$<\delta,\phi>=\phi(0).$$
And now we can answer the question of what $\delta+k$ is if $k$ is a constant. Here we regard $k$ as a constant function; so applying the definitions above gives $$<\delta+k,\phi>=<\delta,\phi>+<k,\phi>=\phi(0)+\int k\phi=\phi(0)+k\int\phi\quad(\phi\in\D).$$
That really is all there is to be said; $<\delta+k,\phi>=\phi(0)=k\int\phi$ really is the definition of $\delta+k$. Honest.
Not the answer you wanted? Sorry, nothing to be done about that. How would we plot $\delta+k$? We don't "plot" $\delta+k$; that makes no sense, because $\delta+k$ is not a function.
David C. Ullrich
- 89,985