 I wanted to give me back the degree of the polynomial in the case when my function is homogenous polynomial. If I do it like that, this will be the radius of the twice ball to the power k. This will be the radius of the ball to the power k, the ratio is two to the power k. So if I want to get the degree, I should take the log, or log base two if you want. It's not so important for us. Say when u is pn, homogenous polynomial of degree n, you'll see that this, and I'll try to write this letter. So it's not the same n, but any more beautiful one of the ball is n log two in my notation. Using their equivalence of norm, we can see that this doubling index, that is bound to it from above. There is a similar estimate from below. The maximum of two b can be bounded by L to norm over four b. So this is to do it more carefully. One over b, and the maximum there is larger than the average. I'm using the local bounded estimate, or equivalence of norm. I know that the supremum of a ball of a solution of elliptic equation is bounded by a constant times L to norm of a larger ball. So there is some constant missing, but we'll add it when we take the logarithms here. And if you look at this ratio, this is my h at four b to the power one-half divided by h at one to the power one-half. You can take the logarithm, you can see that this is bounded by c one, our old function n that was a frequency of four r, c two, where r is the radius of the ball. And in a similar way you can bound it from below and see that this doubling index is comparable to the frequency that we defined before if you allow some change of scale instead of the ball you take four times this ball and you go to estimate from above and they also add the constant there. But if one of those is large, another is large. If doubling is large the frequency is large and was worse. So we'll prefer to work with the doubling index. And then can formulate the claim that is one of the main results for today. Say we are on a compact manifold, we say that there is a constant depending on m such that for any eigenfunction and any relatively small ball the frequency of this eigenfunction or doubling index sorry of the eigenfunction related to this ball is bounded by constant times square root of lambda. Tells us immediately that we can control the vanishing order of eigenfunctions. The eigenfunction with eigenvalue lambda vanishes to the order not larger than square root of lambda. If you compare it to the result we had yesterday when we look at the model lines we know that they are not too dense. So we know that they are relatively dense that there is no ball of radius comparable to one over square root of lambda. This is a very different thing. It tells you that in the center there are not too many nodal lines at least at one point. You don't have intersection of too many nodal lines there. If you think about two dimensional picture the order of vanishing is exactly how many nodal lines you have through one point. So now we have estimate from above on this picture. And to prove it we will use the lifting and general version of three spheres. So first of all go from function phi to function h and let us compare the doubling indices for h and phi. I will start with a ball B in the manifold and take lift of my function lift the manifold and deploy it say by interval minus one one and look at the corresponding ball here. So B prime is a ball with the same center as B same radius but in one more dimension. Then this is the lock of the max over two prime of h over max of h. And I want to estimate the doubling index for eigenfunctions from that one. So I will want to have an estimate from below. The maximum over this ball is larger than the maximum over B prime B pi lambda. What about this one? I take values from here and then multiply them by something and you do it. So the maximum is less than or equal to the maximum over B of phi lambda multiplied by e to the power square root of lambda times r where r is the radius of my ball. And get in this direction. Yes, thank you, you have two B. And then I'll get that this is the doubling of function phi lambda for the ball B minus square root of lambda times r. So the doubling index for the eigenfunction is bounded by the doubling index of its lift or the lifted ball plus term of order square root of lambda. My r will be bounded. So this will be of constant times square root of lambda. So to prove the estimate I need to bound the doubling index of the lifted function. And here we'll use the fact that this is almost monotone and from ball B prime that was a ball here that was probably very small. I can go to a larger one with a fixed radius. And I know that this is bounded by constant times. Lodgeable, all my r's are bounded so the exponential term is bounded by constant here. So we need to estimate the doubling index for ball on a fixed scale. And to do it I want to use the three sphere serum in this form that is still valid for solutions of elliptic PDs. So I will normalize my eigenfunction in a way that the maximum over the manifold of phi lambda is equal to one, just multiplied by a constant. And think about omega as my manifold times minus one one. B is my ball of fixed radius and K will be the part inside. It's a compact subset there. Then three sphere serum for this lifting function gives you that the maximum over this part that is exactly the maximum of the phi times exponential will be one times e to square root of lambda divided by two is bounded by constant maximum of h over this ball to some power times maximum of the whole manifold that is one times the same exponential but present half here, e to square root of lambda one minus gamma. And it tells you that the maximum over the ball is at least constant, e to minus square root of lambda times some constant, thank you. The global maximum is equal to one. The maximum over this ball is at most this one. It means that the doubling of function h over the ball B to primes is bounded by lock of one over this one that is function of this one. So I didn't try this additional constant here but we know that eigenvalues are positive. So it doesn't matter there. So from the lifting and three sphere serum we see that there is a universal bound on the doubling index for eigenfunctions and universal bound for the vanishing order of eigenfunctions. And I think this is all I have time to tell you. Today we'll go back to the doublings on next lecture and see how it works for the result that we want to prove quantitative unique continuation when you replace this ball by a set of positive measure. Thank you very much for your attention today.