Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Say $T$ is the operator defined on class of smooth functions supported on some subset $U$ in $\mathbb{R}^{n}$ where $Tu = -i \sum_{j=1}^{n}{\frac{\partial_{j}\phi}{|\nabla\phi|^2}\partial_{j}u}$ for smooth function $u$ supported on $U$ and $\phi$ some fixed smooth function supported in a neighborhood of $U$.

I want to show that the transpose of $T$ is $^{t}Tu = -Tu + i \frac{\triangle\phi}{|\nabla\phi|^2}u - 2i \sum_{1\leq j,k \leq n}{\frac{\partial_{j}\phi\partial_{k}\phi\partial_{j}\partial_{k}\phi}{|\nabla\phi|^{4}}u}$

Any ideas? I seem to keep getting something else.

share|cite|improve this question
Why that $-i$ constant in front? What have you tried exactly? What goes wrong in that? :) – Patrick Da Silva Aug 4 '13 at 19:04
Um... what matrix? You have an infinite dimensional space here, your operator $T$ cannot be represented as a matrix. Call $V$ your space of smooth functions compactly supported with support $U$ (which I suppose is compact). Then $T : V \to V$ is your linear operator, and $V$ is infinite-dimensional. The transpose has to be computed with respect to some inner product (which I don't know because I never worked with these spaces but it should be standard), and the transpose of $T$, call it $T^{\top}$, must satisfy $$ u \cdot (Tv) = v \cdot (T^{\top} u) $$ for every $u,v \in V$. – Patrick Da Silva Aug 4 '13 at 19:52
I don't think you expect higher order derivatives to appear because the transpose operator is an involution (i.e. $T \circ T : V \to V$ is the identity map, $T(Tu) = u$ for all $u \in V$), so the order of the derivatives taken should remain the same... but that's just a wild guess. – Patrick Da Silva Aug 4 '13 at 19:56
Reading between the lines, is $U$ an open subset of $\mathbb{R}^n$, is your function space the space $C_c^\infty(U)$ of smooth compactly supported complex-valued functions on $U$ with the inner product $$ \langle f,g \rangle = \int_U \overline{f(x)} g(x) d^n x, $$ and by the transpose $^{t}T$ do you mean the adjoint $T^\ast$, i.e., the operator on (presumably) $C_c^\infty(U)$ such that for any $f$, $g \in C_c^\infty(U)$, $$ \langle f, T^\ast g \rangle = \langle T f, g \rangle? $$ Because if so, this is really just a completely standard exercise in integration by parts... – Branimir Ćaćić Aug 4 '13 at 20:28
You guys are right. Sorry. I was not thinking straight. – neelp Aug 4 '13 at 21:42

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.