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bio website math.ucla.edu/~tao
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Professor of Mathematics at UCLA

1d
awarded  Nice Answer
Aug
23
comment Interpolation between $L_p$ and $B^s_{q,q}$
Generally the situation with $p \leq 1$ is quite pathological, particularly the non-convex, non-locally integrable case $p<1$. Things are better if one uses Hardy spaces instead of Lebesgue spaces though.
Aug
22
comment Operator norm vs spectral radius for positive matrices
In fact, your weaker conjecture is true! Let $a_{ij}$ be the $ij$ component of $A$, then $A e_j \geq a_{ij} e_i$ and $A e_i \geq e_j$ using the product partial ordering on ${\bf R}^d$. Iterating, we have $A^{2n} e_j \geq a_{ij}^n e_j$ for any $n$; sending $n$ to infinity we conclude that $a_{ij} \leq \sigma(A)^2$, and so $\|A\| \leq d \sigma(A)^2$ by Schur's test.
Aug
22
comment Operator norm vs spectral radius for positive matrices
Gah, that example doesn't have all entries positive. I don't think that's an essential obstacle, but it does make finding a counterexample trickier. (For instance, the diagonal entries of A are bounded by the trace, which is bounded by d times the spectral radius.) It may be that one has to go to higher dimensions to find a counterexample.
Aug
22
comment Operator norm vs spectral radius for positive matrices
The weaker conjecture is also false; for instance, conjugate $\hbox{diag}(1,2)$ by a large element of $SL_2({\bf Z})$, e.g. $\begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}$. Note again that ill conditioning is the culprit. For integer non-singular matrices one has $|\hbox{det}(A)| \geq 1$, which doesn't give the bounds you want, but may give some other useful inequality for you.
Aug
22
comment Operator norm vs spectral radius for positive matrices
(correction: "positive" should be "positive integer entry", and "positive semi-definite" should be "nonnegative real entry", and "taking limits" should be "rescaling and taking limits" (to first pass from integers to rationals).)
Aug
22
comment Operator norm vs spectral radius for positive matrices
A slightly different way to view Mike's counterexample: if such a bound were true for positive matrices, then by taking limits it would also hold for positive semi-definite matrices, and then a nilpotent matrix would be a counterexample. But this also suggests that one can salvage some bound if one also controlled the condition number of the matrix. Indeed, since the product of the absolute values of the eigenvalues is equal to the product of the singular values (both are equal to $|\hbox{det}(A)|$), one can get some inequality involving the condition number.
Aug
21
comment Small values of a polynomial evaluated at roots of unity
OK, got it. (I got confused and thought $f$ was the minimal polynomial for the sums of roots of unity.) I'm not seeing how the lemma follows from the theorem though - in the limit as $\delta \to 0$, wouldn't one need a very large Lipschitz constant for $\lambda$?
Aug
21
comment Small values of a polynomial evaluated at roots of unity
Why is the Mahler measure of f bounded by k? I thought it was the logarithmic height that was controlled by k, which would make the Mahler measure exponential in k and n.
Aug
21
comment Examples of unproven but likely true existential sentence (in the sense of incompleteness)
Number theory has plenty of unproven existential claims, e.g. (the negation of) en.wikipedia.org/wiki/… .
Aug
20
comment Interpolation between $L_p$ and $B^s_{q,q}$
Ah, I see now. (I had misread the Wikipedia entry.) In that case, one probably has no choice but to use some Calderon-Zygmund theory (as is for instance implicit in the statement that $L^p = F^0_{p2}$) as otherwise one doesn't seem to be able to handle all the varying indices simultaneously.
Aug
20
comment Interpolation between $L_p$ and $B^s_{q,q}$
I don't believe t=p is necessary. Bergh and Lofstrom (for instance) should have details; I don't have access to it here, but the Wikipedia page en.wikipedia.org/wiki/… asserts that the required interpolation result is in Theorem 6.4.5 of that text.
Aug
20
comment Interpolation between $L_p$ and $B^s_{q,q}$
I haven't checked this carefully, but perhaps one could simply use the inclusions $B^0_{p,1} \subset L_p \subset B^0_{p,\infty}$ together with Besov space interpolation?
Aug
18
comment Is it possible to find explicit formula for the product $\prod_{d|n,\ d>1} (1-\mu(d)/\varphi(d))^{\varphi(d)}$?
I think I made a number of sign errors in my previous comments and are no longer able to edit them to correct this, but the broad point of my comments are still valid even if the specific formulae should not be taken literally.
Aug
18
comment Is it possible to find explicit formula for the product $\prod_{d|n,\ d>1} (1-\mu(d)/\varphi(d))^{\varphi(d)}$?
Asymptotics are more interesting; writing $(1 - \frac{\mu(d)}{\phi(d)})^{\phi(d)}$ as $\exp( \mu(d) - \frac{\mu(d)^2}{2\phi(d)} ) (1 + O(\frac{1}{\phi(d)^2} ))$, the product becomes $\exp( -\frac{1}{2} - \sum_{d|n} \frac{\mu(d)^2}{2\phi(d)}) ( 1 + O( \sum_{d|n; d>1} \frac{1}{\phi(d)^2} ))$ which can simplify to $\exp(-\frac{1}{2}-\frac{1}{2}\prod_{p|n} (1+\frac{1}{2(p-1)})) (1 + O( \sum_{p|n} \frac{1}{p^2} )))$. One can be a bit more accurate about the contribution of the small primes $p$ to the error term if one wants more precise asymptotics.
Aug
18
comment Is it possible to find explicit formula for the product $\prod_{d|n,\ d>1} (1-\mu(d)/\varphi(d))^{\varphi(d)}$?
If for instance $n=pq$, then the product is $(\frac{p-2}{p-1})^{p-1} (\frac{q-2}{q-1})^{q-1} (\frac{pq-p-q+2}{(p-1)(q-1)})^{(p-1)(q-1)}$ which has no evident cancellation or further factorisation, and only a small amount of like terms to collect. Doesn't seem like there is much else to be done here.
Aug
18
awarded  Populist
Aug
17
comment Is there a generalization of Sobolev spaces for certain locally compact groups?
There is some literature on Sobolev spaces in arbitrary metric measure spaces: www2.pitt.edu/~hajlasz/OriginalPublications/… . For a locally compact group, one can use Haar measure for the measure, and if the group is second countable one can use Birkhoff-Kakutani to get a metric (but the choice of metric is not unique, and this can lead to different Sobolev spaces, e.g. a Riemannian metric gives different results to a Carnot-Caratheodory metric).
Aug
14
comment For integers $a \ge b > 1$ is $f(a,b) = a^b + b^a$ injective?
The problem looks at least as hard as the Catalan conjecture en.wikipedia.org/wiki/Catalan's_conjecture , which concerns the equation $a^b = c^d + 1$ instead of $a^b + b^a = c^d + d^c$, and an unconditional solution is likely out of reach of current methods. The 4-variable version of the abc conjecture (see jstor.org/stable/2153551 ) may be able to imply the conjecture (or at least a large fraction of it), though.
Aug
13
comment Geometric Interpretation of Trace
This property, together with linearity, determines the trace uniquely, and so one can view the trace as the linearised version of the dimension-counting operator. (This is basically the "noncommutative probability" way of thinking about the trace.)