1
$\begingroup$

Let $L: C^k(U) \rightarrow C(U)$ be a linear differential operator on the open set $U \subset \Bbb R^n$ that possesses the order $k \in \Bbb N_0$. Show that $(L^*)^* = L$.

Please note: I'd appreciate to receive only a tiny hint. :-)

My attempt so far:

We are allowed to assume that $L^*$ is the adjoint differential operator of $L$ and that $(L^*)^*$ is the adjoint differential operator of $L^*$. These adjoint operators are unique. Since $L^*$ is adjoint to $L$ and $(L^*)^*$ is adjoint to $L^*$, we know that

$\int_U (L^* f)gd^n x = \int_U f(Lg)d^nx$

and

$\int_U ((L^*)^* f) gd^n x = \int_U f(L^*g)d^nx$.

Now, what the theorem actually states is:

$L$ is the adjoint differential operator of $L^*$.

Therefore, it is reasonable to assume that I have to show:

$\int_U (Lf) g d^n x = \int_U f(L^*g) d^nx.$

Yet, I have failed to find a way to go from here.

Edit:

I just don't get it. Does anyone have another tyni hint that builds up on the answer I already received?

  • 0
    Can you let us know what definition of adjoint you are using, with the spaces in which the functions live? – 2017-01-28
  • 0
    I am using this one: https://en.wikipedia.org/wiki/Differential_operator#Adjoint_of_an_operator – 2017-01-28

2 Answers 2

2

You seem very close, because you want to show

$$\int_U (Lf) g \,d x = \int_U f(L^*g) \, dx,$$ and you know that $L^*$ is the adjoint of $L$.

  • 0
    Let me think about that for a moment. – 2017-01-21
  • 0
    I just don't get it. – 2017-01-24
  • 0
    Hmm, I am not sure to understand what exactly you don't understand. Since $L$ is a bounded linear operator between the Banach spaces $C^k(U)$ and $C(U)$, the adjoint is strictly speaking from $C(U)^*$ to $C^k(U)^*$. (see https://en.wikipedia.org/wiki/Hermitian_adjoint) But since you wrote the formula with the integrals, I assumed that you were referring to the formal adjoint (https://en.wikipedia.org/wiki/Differential_operator#Adjoint_of_an_operator). Can you write the definition of the adjoint that you are using? – 2017-01-25
0

If your functions are valued in a field (i. e. $\mathbb{R}$) it is safe to assume that

$$ \int_U fg \, \mathrm{d}^n x = \int_U gf \, \mathrm{d}^n x $$

for all $f,g \in C(U)$.

  • 0
    But it's not true that $\int_U (L*f)g = \int_U (L*g)f$, is it? – 2017-01-24
  • 0
    No, it's not true. You are correct. If $(L * f) = L^* f$. – 2017-01-24
  • 0
    I don't see how to apply your hint then. :( – 2017-01-24
  • 0
    you can substitute $f$ with $(Lf)$ for the start. when apply known identities. – 2017-01-24
  • 0
    So you want me to write $\int_U (Lf)g = \int_U fg = \int_U gf = \int_U g(Lf)$? – 2017-01-24
  • 0
    Sorry for abuse of notation. more like $ \int_U(Lf)g = \int_U hg = \int_u gh = \int_U g(L f) $with $h = (Lf) \in C(U)$. – 2017-01-24
  • 0
    Oh, okay, I'll give it a try. :-) – 2017-01-24
  • 0
    Basically, $\int_U (Lf)g = \int_U g(Lf) = \int_U (L^*g)f = \int_U f(L^*g)$? – 2017-01-24
  • 0
    This doesn't make sense though. It's fine for $g$ to be the outer function here, but in order to apply the definiton of $L^*$, we need $f \in C_c^k$. This is, of course, not true in general. – 2017-01-24
  • 0
    Yes, this what I was trying to say. Note, that $f \in \mathrm{Dom} \, L = C^k(U)$ and $g \in \mathrm{Dom} L^* = C(U)^*$, having $C_c^k(U) \subset C(U)^*$ but $C_c^k(U) \neq C(U)^*$. However, definition of adjoint operator must still apply in this case if you interchange $f$ and $g$ in the definition in the way that preserves the domains. – 2017-01-25