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June 28

Everywhere discontinuously differentiable function

Can someone please give, if possible, an example of a (real-valued) function that is everywhere differentiable, but whose derivative is everywhere discontinuous? If such a function can't exist, why?--Jasper Deng (talk) 10:40, 28 June 2015 (UTC)[reply]

Apparently not... — 79.113.201.52 (talk) 14:03, 28 June 2015 (UTC)[reply]

If you want some fun "counterexamples", get hold at Counterexamples in Analysis by Gelbaum and Olmsted. It doesn't give this particular one, but many others. YohanN7 (talk) 10:12, 1 July 2015 (UTC)[reply]

The set of discontinuities of a derivative is a set of the first category. In particular, by the Baire category theorem, its complement is non-empty. For proof, see John Oxtoby, "Measure and category". Sławomir Biały (talk) 11:03, 1 July 2015 (UTC)[reply]

For a set in [0, 1] of category I and measure 1 and a set in [0, 1] of measure 0 and category II, see above mentioned book. YohanN7 (talk) 11:49, 1 July 2015 (UTC)[reply]

Almost conjugate matrices

We know that there is a nonsingular matrix A so AX=YA iff X and Y are conjugate iff they have the same Jordan normal form (at least over a complete field). What about when AX=YA where A is singular but still nonzero? The set {A:AX=YA} is a vector space which can have dimension from 0 to n² depending on X and Y. If X and Y are diagonalizable and they have no eigenvalues in common then the dimension is 0, but what can be said about the dimension if they do have eigenvalues in common? And what about the nondiagonalizable case? I was mainly interested in 2×2 rotation matrices X = R(α), Y = R(β). In this case the dimension is 4 if α=β=0 or α=β=π; 2 if α=±β; and 0 otherwise. But I thought the more general question might be interesting as well. --RDBury (talk) 20:41, 28 June 2015 (UTC)[reply]