bio | website | regularize.wordpress.com |
---|---|---|
location | Braunschweig, Germany | |
age | 36 | |
visits | member for | 4 years, 11 months |
seen | 3 hours ago | |
stats | profile views | 2,005 |
Professor at TU Braunschweig.
Area: Inverse problems, regularization theory, applied functional analysis, mathematical image processing.
Apr
14 |
comment |
2, 3, and 4 (a possible fixed point result ?)
As far as I understood, the BGK (Browder-Göhde/Göbel-Kirk) fixed point theorem states that every non-expansive self-mapping on a non-empty, closed and convex subset of a uniformly convex Banach space has a fixed point. |
Apr
7 |
awarded | Necromancer |
Mar
27 |
comment |
Hadamard / matrix product adjoint
Ok, I see now. I think a better fit for this question would be scicomp.stackexchange.com. Anyway: The adjoint is defined by $\langle A^* y,x\rangle = \langle y,Ax\rangle$. You do adjoints one by one from outside to inside. |
Mar
26 |
comment |
Hadamard / matrix product adjoint
Sorry, but this does not make sense. What is the Hadamard product of the matrix S and the vector Fx? |
Mar
26 |
comment |
Hadamard / matrix product adjoint
Err... So x is a matrix? |
Mar
26 |
comment |
Hadamard / matrix product adjoint
I could not parse your definition of the operator A. If D and S are matrices, do you build the Hadamard product with the matrix of the Fourier transform? |
Mar
26 |
comment |
Convergence in energy of bounded (semi)subharmonic functions
Sorry for the late reply: I guess the best thing would be to write an answer yourself so that this is kept for the records. |
Mar
26 |
reviewed | Approve Are all rational exactly solvable differential equations known? |
Mar
24 |
awarded | Nice Answer |
Mar
24 |
reviewed | Leave Open Which way for reading the proofs? |
Mar
24 |
answered | Which way for reading the proofs? |
Mar
23 |
reviewed | Leave Closed If a polynomial $p(z)$ omits a value, then $p(z)-\dfrac{(1-e^{i\psi})}{n}zp^{\prime}(z)$ also omits that value |
Mar
23 |
reviewed | Leave Open Analysis of Sobolev spaces |
Mar
17 |
comment |
Fredholm integral with functions constrained to [0;1]
For some reason I confused $g$ and $f$. Of course $f$ should be the unknown… Corrected. |
Mar
17 |
revised |
Fredholm integral with functions constrained to [0;1]
corrected variable names |
Mar
16 |
comment |
Fredholm integral with functions constrained to [0;1]
The most simple thing that comes to mind is the projected gradient method, see my update. |
Mar
16 |
revised |
Fredholm integral with functions constrained to [0;1]
added 361 characters in body |
Mar
16 |
answered | Fredholm integral with functions constrained to [0;1] |
Mar
10 |
reviewed | Leave Open More recently published comprehensive reference on inequalities in the spirit of Hardy-Littlewood-Pólya |
Mar
6 |
awarded | Necromancer |