# Proof of uniqueness about distribution in Rudin's

I'm reading Functional Analysis by Rudin, and have trouble understanding a part of the proof of theorem 6.33, in page 174. This theorem states an one-to-one relationship between a linear continuous mapping $L:\phi\mapsto u*\phi$ from $\mathscr{D}$ to $\mathscr{C}^{\infty}$ and a distribution $u\in \mathscr{D}'$.

At the end of proof, it tries to confirm the uniqueness of $u$. It says

The uniquess of $u$ is obvious, for if $u\in\mathscr{D'}$ and $u*\phi=0$ for every $\phi\in\mathscr{D}$, then $u(\check{\phi})=(u*\phi)(0)=0$ for every $\phi\in \mathscr{D}$; hence $u=0$.

I don't quite understand why this is obvious...Actually I don't see why the "for" part implies the uniqueness.

• I'd guess that $u \mapsto (\phi \mapsto u * \phi)$ is linear so he just says that the kernel of that map is $0$, hence proving the injectivity of that map (and so the uniqueness of $u$). Commented Mar 16, 2016 at 12:12
• @xavierm02 Oh, I guess I see your point. Thanks!
– Hua
Commented Mar 16, 2016 at 12:49
• @xavierm02 Do you mind changing this comment to an answer post? I think this comment is clearer to understand the reasoning behind the proof.
– Hua
Commented Mar 17, 2016 at 9:38
• did you understand why $u(\tilde{\phi}) = (u \ast \phi) (0)$?? this is part of the definition of the convolution Commented May 28, 2016 at 8:26
• @user1952009 Yes, I know this is the definition of convolution.
– Hua
Commented May 28, 2016 at 8:28

The key is that linear functions on vector spaces are one-to-one iff they only map $0$ to $0$, or said another way, iff the kernel contains exactly 0.

For the forward direction, suppose $L:X\to Y$ is not one-to-one. Then $\exists$ $x \ne y$ such that $L(x) = L(y)$. By linearity, this implies $L(x) - L(y) = L(x - y) = 0.$ Therefore the kernel of $L$ contains $x - y \ne 0.$

For the other direction, if the kernel contains a vector $x \ne 0$ such that $L(x) = 0$, then trivially $L(x) = 0$ and $L(0) = 0$, so the mapping is not one-to-one.

• I understand this property of linearity. But for my question, we need to identify the linear mapping first. Right?
– Hua
Commented May 28, 2016 at 8:29
• In a comment by @xavierm02, he gives a linear mapping which I agree with. But the answer by Jonas says that 'the uniqueness of the zero distribution implies the uniqueness of all others', which I don't understand. I don't think only the linearity of distributions can imply the uniqueness in my question.
– Hua
Commented May 28, 2016 at 8:37

Distributions are linear functionals, and so $u=v$ is equivalent to $u-v=0$. Thus, the uniqueness of the zero distribution implies the uniqueness of all others.

• I guess I can get what you are trying to say. But do you mind expanding your answer, so that it is clearer enough to be accepted?
– Hua
Commented Mar 16, 2016 at 12:54
• Sorry, it is quite clear. I really don't care whether it is accepted or not. Commented Mar 16, 2016 at 13:54
• OK...actually I'm thinking if you misunderstand my question in which an one-to-one corresponding is being established between distributions and a set of mappings. So where does the linearity of $u$ itself come in? I guess the zero distribution you mentioned in your answer is the $0$ element in the space $\mathscr{D'}$. But the $0$ element in a vector space is unique no matter what space it is. So I don't understand why linearity of $u$ implies the uniquess of all others (actually what others mean here?)
– Hua
Commented Mar 16, 2016 at 14:22
• Good luck! ---- Commented Mar 16, 2016 at 18:43
• I just wana every question I asked and answers I got will be helpful and understandable to the others, though some questions may be silly or stupid. Rudin's book is popular but dense and lack of motivation. I believe many self-learners meet many problems as I do.
– Hua
Commented Mar 17, 2016 at 9:46

I here expand the comment by @xavierm02. Credits go to him.

Define a mapping $T: u\mapsto(L: \phi\mapsto u*\phi)$. Clearly $T$ is linear. The zero element in the codomain is the $L_0$ with a $u_0$ in $\mathscr{D}'$ such that $L_0(\phi)=u_0*\phi=0$ for any $\phi$ in $\mathscr{D}$. Here $0$ is constant function. Then $u_0(\check{\phi})=(u_0*\phi)(0)=0$ for any $\phi$ in $\mathscr{D}'$. The first equal sign is by the definition of the operation $*$. In other words, $u_0$ is the zero element in $\mathscr{D}'$ ,which is unique because of the fact that $\mathscr{D}'$ is a linear space and a linear space has a unique zero element. This means $kern(T)=\{0\}$, which implies $T$ is one-to-one and in turn guarantees the uniqueness of $u$ for a $L$.

For the answer by @Jonas, though his conclusion is right, I don't think the reasoning behind the first sentence "Distributions are linear functionals, and so u=vu=v is equivalent to u−v=0u−v=0." is right. The uniqueness of zero element of a vector space comes from the nature of a vector space, not from the linearity of the elements, i.e., the functionals in this case, in the space.