bio | website | regularize.wordpress.com |
---|---|---|
location | Braunschweig, Germany | |
age | 36 | |
visits | member for | 4 years, 11 months |
seen | yesterday | |
stats | profile views | 2,005 |
Professor at TU Braunschweig.
Area: Inverse problems, regularization theory, applied functional analysis, mathematical image processing.
Jan
8 |
comment |
Elaborating Mercer's theorem (RKHS) on Cameron-Martin space $k(x,y)=\min(x,y)$
Seems like I did not look careful enough on my mesh plot... Sorry. |
Jan
8 |
comment |
Elaborating Mercer's theorem (RKHS) on Cameron-Martin space $k(x,y)=\min(x,y)$
Seems to be a bug in your numerics... Calculations look fine and my plots also... |
Jan
8 |
answered | Smoothing L1 norm, Huber vs Conjugate |
Jan
8 |
comment |
Smoothing L1 norm, Huber vs Conjugate
Smoothing the dual will not give you a smooth primal. However, you get a strongly convex primal by dual smoothing (as opposed to merely a strictly convex primal by Huber smoothing). Hence, it depends on what kind of regularity you are aiming at: A smoother primal or a "more convex" primal - both can be helpful algorithmically. Moreover, note that there are numerous methods to treat nonsmooth convex minimization problems efficiently. |
Jan
5 |
revised |
New grand projects in contemporary math
Extended description. |
Jan
2 |
answered | New grand projects in contemporary math |
Dec
17 |
comment |
Most memorable titles
Reminds my of books with titles like "Theory of normal families". I've been told that one of these could be found in the "social sciences" section the university library in Bremen... |
Dec
9 |
comment |
How to solve a system of linear equations without storing the matrix?
Granted, convergence can be slow (in terms of iteration count and computational effort - its advantage is low memory). You don't need any requirements for the matrix (for the complex case adjust the projection accordingly). In fact you could also apply the method to rectangular systems. It converges to some solution for the underdetermined case (and the minimum-norm solution if initialized with zero). In the overdetermined case you need to stop at some point as you'll see that the residual $\|AX-Y\|$ is not decreasing anymore. |
Dec
7 |
answered | How to solve a system of linear equations without storing the matrix? |
Dec
7 |
answered | If d/dx is an operator, on what does it operate? |
Dec
5 |
answered | ODE's without a Lipschitz condition |
Nov
24 |
comment |
fixed point arguments in PDE
Would Schauder's fixed point theorem be standard or advanced? |
Nov
14 |
comment |
Why is the Gaussian so pervasive in mathematics?
Isn't this connection the reason why the Gaussian is called Gaussian? If I remember correctly, Gauss introduced "least squares" as a regression technique not because its derivative gives rise to a linear problem but because the square fits the "normal error model", i.e. corresponds to the Gaussian distribution. |
Nov
13 |
comment |
Weak convergence of the image of a weakly $L^1$ converging sequence
Not an answer, but I vaguely recall that the only functions $f$ for which $u_k$ weakly to $u$ implies $f(u_k)$ weakly to $f(u)$ are affine linear $f$'s. |
Nov
5 |
awarded | Nice Answer |
Nov
1 |
answered | A sudden smiley? :-) |
Oct
22 |
awarded | Popular Question |
Oct
20 |
comment |
Inversion of Radon transform by incomplete data: specific case
Could you expand on the relation of the Radon transform and economics? (Out of curiosity.) |
Oct
19 |
comment |
What software one needs to solve a big linear system on a small computer?
I also recommend octave. 10000 x 10000 is doable, although the matrix needs 800MB (if not sparse). On my laptop it takes about a minute to solve a system like this. |
Oct
16 |
comment |
Measuring almost-critical values of smooth functions.
Did you inspect the one-dimensional case $X=\mathbb{R}$? I suspect that basically arbitrary functions $M_f$ are possible. |