1
$\begingroup$

In a proof for the fact that every distribution has an antiderivative (in Introduction to Hilbert space with application(Debnath,Mikusinski)) they use the fact that for any $$ \phi_0 \in D(R) $$ s.t $$ \int_{-\infty}^{\infty}\phi_0(x)dx = 1$$ we can represent every test function $$ \phi \in D(R), \space \space \phi = K\phi_0 + \phi_1$$ where $$ K = \int_{-\infty}^{\infty}\phi(x)dx, \space \int_{-\infty}^{\infty}\phi_1(x)dx = 0$$ now I'm stuck on an exercise that wants me to prove the representation is possible for every phi aswell as the fact that the represention is unique, I've tried to reach some equality between the two when I apply a distribution but my understanding of distributions and test function is very limited aswell as my intuition for these kinds of problems.

I'm having trouble getting started in the right direction so any help is appreciated.

$\endgroup$
0
$\begingroup$

Fix $\phi_0 \in \mathcal{D}(\mathbb R)$ such that $\int_{-\infty}^{\infty} \phi_0(x) \, dx = 1$.

For an arbitrary $\phi \in \mathcal{D}(\mathbb R)$ let $K = \int_{-\infty}^{\infty} \phi(x) \, dx$ and $\phi_1 = \phi - K\phi_0.$

Show that $\phi_1 \in \mathcal{D}(\mathbb R)$. This shows that "the representation is possible for every phi".

Assume that there is another represention $\phi = L\phi_0 + \phi_2,$ where $L = \int_{-\infty}^{\infty} \phi(x) \, dx$ and $\int_{-\infty}^{\infty} \phi_2(x) \, dx = 0.$ Show that $K=L$ and $\phi_1=\phi_2.$ This proves uniqueness.

$\endgroup$
5
  • $\begingroup$ I'm guessing you mean $$ \int_{-\infty}^{\infty}\phi_0(x)dx = 1 $$ and arbitrary $$\phi \in \mathcal{D}(\mathbb R) $$ and to show that it is in D would I then want to prove it is smooth and vanishing? $\endgroup$ Mar 21 '20 at 21:59
  • $\begingroup$ Of course I meant that. I have edited my answer. $\endgroup$
    – md2perpe
    Mar 22 '20 at 7:17
  • $\begingroup$ Well, you hardly need to show that $\phi_1$ is smooth and vanishing. You know that $\mathcal{D}(\mathbb R)$ is a linear space, don't you? $\endgroup$
    – md2perpe
    Mar 22 '20 at 7:18
  • $\begingroup$ Oh ok that's quite simple then if I'm understanding correctly, since $\phi_1 $ is a linear combination of test functions we say $ \phi_1 \in \mathcal{D}(\mathbb R) $ which means it's possible for every phi. I'm sorry if I'm a bit slow with this, as I said earlier I have very limited experience with test functions and sobolev spaces and the book only gives a brief overview. $\endgroup$ Mar 22 '20 at 16:25
  • $\begingroup$ You have understood correctly. $\endgroup$
    – md2perpe
    Mar 22 '20 at 20:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.