# Dirac delta function + constant [closed]

If we sum the Dirac delta function with a constant, what is the result? I.e., $$k+\delta(x)$$, where $$k$$ is a constant.

• Think about the values that the Delta function takes on over the real line and at $x=0$. – aleden Jan 7 '19 at 13:30
• something like: $+\infty$ when $x=0$ and $0$ in everywhere else? – ARF Jan 7 '19 at 13:40
• The 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 7 '19 at 13:42
• @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 7 '19 at 14:23
• @PeterMelech The pairing is linear : $<k\delta,\phi>=k\phi(0)$. – David C. Ullrich Jan 7 '19 at 14:54

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 $$:=u(\phi)$$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 $$=\int f\phi.$$(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 $$=+$$and $$=k.$$

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 $$<\delta+k,\phi>=<\delta,\phi>+=\phi(0)+\int k\phi=\phi(0)+k\int\phi\quad(\phi\in\D).$$

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.