Moderator Notice: I am unilaterally closing this question for three reasons.

- The discussion here has turned too chatty and not suitable for the MSE framework.
- Given the recent pre-print of T. Tao (see also the blog-post here), the continued usefulness of this question is diminished.
- The final update on this answer is probably as close to an “answer” an we can expect.
Eminent Kazakh mathematician

Mukhtarbay Otelbaev, Prof. Dr. has published a full proof of the Clay Navier-Stokes Millennium Problem.Is it correct?

See http://bnews.kz/en/news/post/180213/

A link to the paper (in Russian):

http://www.math.kz/images/journal/2013-4/Otelbaev_N-S_21_12_2013.pdfMukhtarbay Otelbaev has published over 200 papers, had over 70 PhD students, and he is a member of the Kazak Academy of Sciences. He has published papers on Navier-Stokes and Functional Analysis.

please confine answers to any actual mathematical error found!

thanks

**Answer**

This web page has Theorem 6.1. It is written in Spanish, but actually is rather easy to follow even if (like me) you don’t know any Spanish. However it is not made clear on this web site that the statement of Theorem 6.1 is “**If** $\|A^\theta \overset 0u\| \le C_\theta\|$, **then** $\| \overset0u \| \le C_1(1+\|\overset0f\|+\|\overset0f\|^l)$.”

This is the proposed counterexample to Theorem 6.1 given at http://dxdy.ru/topic80156-60.html. I used google translate, and then cleaned it up. I also added details here and there.

Let $\hat H = \ell_2$.

Let the operator $A$ be defined by $ Ae_i = e_i $ for $ i < 50$, $ Ae_i = ie_i $ for $ i \ge $ 50

Define the bilinear operator $ L $ to be nonzero only on two-dimensional subspaces $ L (e_ {2n}, e_ {2n +1}) = \frac1n (e_ {2n} + e_ {2n +1}) $, with $ n \ge 25 $.

Check conditions:

U3. Even with a margin of 50 .

U2. $ (e_i, L (e_i, e_i)) = 0 $ for $ i \ge $ 50. This is also true for eigenvectors $ u $ with $ \lambda = 1 $, because for them $ L (u, u) = 0 $.

U4. $ L (e, u) = 0 $ for the eigenvectors $ e $ with $ \lambda = 1 $ also trivial. (Stephen’s note: he also needs to check $L_e^*u = L_u^*e = 0$, but that looks correct to me.)

U1. $ (Ax, x) \ge (x, x) $. Also $$ \| L (u, v) \| ^ 2 = \sum_{n \ge 25} u ^ 2_ {2n} v ^ 2_ {2n +1} / n ^ 2 \le C\|(u_n/\sqrt n)\|_4^2 \|(v_n/\sqrt n)\|_4^2 \le C\|(u_n/\sqrt n)\|_2^2 \|(v_n/\sqrt n)\|_2^2 = C \left (\sum u ^ 2_ {n} / n \right) \left (\sum v ^ 2_ {n} / n \right) $$ so we can take $\beta = -1/2$.

And now consider the elements $ u_n =-n (e_ {2n} + e_ {2n +1}) $. Their norms are obviously rising.

Let $ \theta = -1 $. Then the $A^\theta $-norms of all these elements are constant. But, $ f_n = u_n + L (u_n, u_n) = 0 $.

**Update:** Later on in http://dxdy.ru/topic80156-90.html there is a response relayed from Otelbaev in which he asserts he can fix the counterexample by adding another hypothesis to Theorem 6.1, namely the existence of operators $P_N$ converging strongly to the identity such that one has good solvability properties for $u + P_N L(P_N u,P_N u) = f$, in that if

$\| f \|$ is small enough then $\| u \|$ is also small.

Terry Tao communicated to me that he thinks a small modification of the counterexample also defeats this additional hypothesis.

**Update 2:** Terry Tao modified his example to correct for that fact that the statement of Theorem 6.1 is that a bound on $u \equiv \overset0u$ implies a lower bound on $f \equiv \overset0f$ rather than the other way around (i.e. we had a translation error for Theorem 6.1 that I point out above).

Let $\hat H$ be $N$-dimensional Euclidean space, with $N \ge 50$. Let $\theta = -1$ and $\beta = -1/100$. Take

$$ A e_n = \begin{cases} e_n & \text{for $n<50$} \\

50\ 2^{n-50} e_n & \text{for $50 \le n \le N$.}\end{cases}$$

and

$$L(e_n, e_n) = – 2^{-(n-1)/2} e_{n+1} \quad\text{for $50 \le n < N$,}$$

and all other $L(e_i,e_j)$ zero.

Axioms (Y.2) and (Y.4) are easily verified. For (Y.1), observe that

if $u = \sum_n c_n e_n$ and $v = \sum_n d_n e_n$, then for a universal constant $C$, we have

$$ \| L(u,v) \|^2 \le C \sum_n 2^{-n} c_n^2 d_n^2, \\

|c_n| \le C 2^{n/100} \| A^\beta u \| ,\\

|d_n| \le C 2^{n/100} \| A^\beta v \| ,$$

and the claim (Y.1) follows from summing geometric series.

Finally, set

$$

u = \sum_{n=50}^N 2^{n/2} e_n

$$

then one calculates that

$$

\| A^\theta u \| < C

$$

for an absolute constant C, and

$$

u + L(u,u) = 2^{50/2} e_{50}

$$

so

$$

\| u + L(u,u) \| \le C

$$

but that

$$

\| u \| \ge 2^{N/2}.

$$

Since $N$ is arbitrary, this gives a counterexample to Theorem 6.1.

By writing the equation $u+L(u,u)=f$ in coordinates we obtain $f_n = u_n$ for $n \le 50$, and $f_n = u_n + 2^{-n/2} u_{n-1}^2$ if $50<n\le N$. Hence we

see that $u$ is

uniquely determined by $f$. From the inverse function theorem we see

that if $\| f \|$ is sufficiently small then $\| u \| < 1/2$, so the

additional axiom Otelbaev gives to try to fix Theorem 6.1 is also

obeyed (setting $P_N$ to be the identity).

**Update 3:** on Feb 14, 2014, Professor Otelbaev sent me this message, which I am posting with his permission:

Dear Prof. Montgomery-Smith,

To my shame, on the page 56 the inequality (6.34) is incorrect therefore the proposition 6.3 (p. 54) isn’t proved. I am so sorry.

Thanks for goodwill.

Defects I hope to correct in English version of the article.

**Attribution***Source : Link , Question Author : Community , Answer Author :
14 revs*