1. Aug 11, 2014

### joshmccraney

hey pf!

can you help me understand what an adjoint operator is? i've read lots of threads and other sites, but am having trouble. maybe you could give me an example?

for example, does the operator d/dx have an adjoint? is asking this question completely stupid of me?

thanks!

2. Aug 12, 2014

### johnqwertyful

First of all you need to have an inner product space (there are other ways to define it, in my functional analysis class we defined adjoints on general Banach spaces. But don't worry about that). We generally assume that the inner product space is "complete" or Hilbert. Don't worry about it now if you don't know. We will just consider inner product spaces.

An inner product space is a vector space, $V$ over $\mathbb{C}$ (or $\mathbb{R}$) with a function defined on it.
$$(,)\colon V \times V \rightarrow \mathbb{C}$$
(or $\mathbb{R}$)

such that:
$(v,v)\geq 0$ for all $v \in V$
$(v,v)=0$ iff $v=0$
$(u,av+bw)=a(u,v)+b(u,w)$ for all $a,b\in \mathbb{C}$ and $u,v \in V$
$(u,v)=\overline{(v,u)}$ for all $u,v\in V$

If you have a linear map, $A\colon V\rightarrow V$, the adjoint $A^{\ast}$ is a linear map such that:
$(u,Av)=(A^{\ast}u,v)$ for all $u,v\in V$

Basically, the adjoint is what allows you to move an operator from one side of an inner product to the other. So in your question, what is the adjoint of $\frac{d}{dx}$? First of all you need an inner product space. Let $V=C^{\infty}_{c}(\mathbb{R})$ be the vector space of all compactly supported smooth functions (for simplicity, lets consider only real functions). You can check that this is a vector space (proof is LTR =) ), let the inner product be:
$$(f,g)=\int_{-\infty}^{\infty} f(x) g(x) dx$$
So
$$(f,\frac{dg}{dx})=\int_{-\infty}^{\infty} f(x)\frac{dg}{dx}(x) dx$$
We may integrate by parts
$$(f,\frac{dg}{dx})= f(x)g(x)|_{-\infty}^{\infty}-\int_{-\infty}^{\infty} \frac{df}{dx}(x) g(x)dx$$
Since the functions are compactly supported, end terms disappear, so
$$(f,\frac{dg}{dx})= -\int_{-\infty}^{\infty} \frac{df}{dx}(x) g(x)dx$$
Therefore
$$(f,\frac{dg}{dx})=(-\frac{df}{dx},g)$$
So on this space, the adjoint of $\frac{d}{dx}$ is $-\frac{d}{dx}$.
However this is just on this space! If we assumed that there are boundary terms, it gets more complicated, if we are over complex numbers it gets more complicated. If functions aren't smooth, or even differentiable, it gets way more complicated.

Last edited: Aug 12, 2014
3. Aug 12, 2014

### joshmccraney

thanks! i really appreciate that! i have a much better idea what is going on. could you help me on a very remaining points?

sorry, but what does this notation mean and what do you mean my "compactly supported smooth functions"?

maybe i'll know why once i understand compactly supported, but why is $f(x)g(x)|_{-\infty}^{\infty}=0$?

thanks again!

4. Aug 12, 2014

### johnqwertyful

Compactly supported means that the function is 0 outside of some interval.
Smooth means infinitely differentiable.

5. Aug 12, 2014

### joshmccraney

i see. does this relate to closed and bounded at all, or are you speaking of some other compact definition?

6. Aug 12, 2014

### johnqwertyful

Compactly supported means the function is 0 outside of some compact set.

Last edited: Aug 12, 2014
7. Aug 12, 2014

### johnqwertyful

Also compact means that every open cover has a finite subcover. "closed and bounded" is not a definition, but a theorem. A set of real numbers is compact iff it is closed and bounded.

8. Aug 12, 2014

### joshmccraney

Yea, the HB theorem I think (it's been a while since analysis). Weird, why do they call it compactly supported? (I know that's more of a history buff)

9. Aug 12, 2014

### johnqwertyful

http://en.wikipedia.org/wiki/Support_(mathematics [Broken])

The support of a function is where it doesn't vanish. I was being loose at first, because I didn't think you knew what compact meant. Compactly supported means that it vanishes on all but a compact set. I said "vanishes outside of an interval" because if the function vanishes outside of a compact set, there is an interval around the compact set.

None of this matters to the point I was making. The point was that the end terms vanish.

Last edited by a moderator: May 6, 2017
10. Aug 12, 2014

### WWGD

Think of the Riesz rep. theorem as related/motivation. Maybe a good exercise to show that
in R^n, if the map is described by a matrix A, then the adjoint is A^T, the transpose. Note that in the infinite-dimensional case, the adjoint may not exist.

11. Aug 13, 2014

### Fredrik

Staff Emeritus
I'm using the physicist's convention for inner products, so my inner product is linear in the second variable and antilinear (=conjugate linear) in the first.

Let H be a Hilbert space over $\mathbb C$. For each $x\in H$, I will use the notation $\langle x,\cdot\rangle$ for the map $y\mapsto \langle x,y\rangle$ with domain H. The Riesz representation theorem tells us that for each $\phi\in H^*$ (where H* is the set of bounded linear maps from $H$ into $\mathbb C$), there's a unique $x\in H$ such that $\phi=\langle x,\cdot\rangle$.

Let A be a bounded linear operator on H. For each $x\in H$, the map $y\mapsto\langle x,Ay\rangle$ with domain A is linear and bounded. (It's elementary to show linearity. Boundedness follows from the Cauchy-Schwartz inequality). Let's denote this map by $\phi_{x,A}$. The Riesz representation theorem tells us that there's a unique $z\in H$ such that $\phi_{x,A}=\langle z,\cdot\rangle$. The map $x\mapsto z$ with domain H is what we call the adjoint of A. It's denoted by $A^*$ or $A^\dagger$, so we can write $A^*x=z$.

What I said in the preceding paragraph implies that for all $x,y\in H$, we have
$$\langle x,Ay\rangle =\phi_{x,A}(y) =\langle z,y\rangle =\langle A^*x,y\rangle.$$
This isn't super easy, but it can't be made much easier than this if you want it done rigorously.

The definition, and proof of existence, for not necessarily bounded operators (like d/dx) is similar to the above, but it's more complicated, because we have to pay attention to the domains of both $A$ and $A^*$.