Timeline for The missing link: an inequality
Current License: CC BY-SA 3.0
11 events
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| Jan 13, 2017 at 21:57 | history | edited | Dima Pasechnik | CC BY-SA 3.0 |
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| Jan 13, 2017 at 17:11 | comment | added | Iosif Pinelis | @DimaPasechnik : Of course, except for the case $x=0+$, the condition $y=0+$ means $n\to\infty$. I had this in mind, but I haven't looked closely at the case $n\to\infty$. | |
| Jan 13, 2017 at 16:05 | history | edited | Dima Pasechnik | CC BY-SA 3.0 |
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| Jan 13, 2017 at 16:02 | comment | added | Dima Pasechnik | IMHO $y=0+$ means that we are in the asymptotic case $n>> 1$, and this is apparently possible to solve in the affirmative by different means, see the answer by Matt Young. (But certainly, thanks, I stand corrected) | |
| Jan 13, 2017 at 15:27 | comment | added | Iosif Pinelis | The reason for $F$ not to be convex, even in the weaker sense, is that $F_{xx}=0$ if $y=0+$, whereas $F_{yx}=-2(1-x)\ne0$ if $y=0+$. The case $y=0+$ seems the most difficult in the somewhat similar approach that I suggested. | |
| Jan 13, 2017 at 13:59 | comment | added | Iosif Pinelis | @DimaPasechnik : Alas, your function $F$ is not convex even for $y<x$, since the determinant of the Hessian matrix is $-4(1-x)^2<0$ at $y=0+$. In fact, something less than the convexity of $F$ would have sufficed (in addition to $F_y\ge0$). The subscripts here denote the differentiation in the corresponding variables. Indeed, $F''_n(x)=F_y y_{xx}+F_{xx}+F_{yx}y_x+F_{yy}y_x^2$. So, it would be enough that $F_y\ge0$, $F_{xx}\ge0$, and the discriminant $d:=F_{yx}^2-4F_{xx}F_{yy}$ be $\le0$. However, $d=4(1-x)^2>0$ if $0<x<1$ and $y=0+$. | |
| Jan 13, 2017 at 13:44 | comment | added | user44143 | Tarski's results tell us that something like this actually follows from the original claim. The claim tells us that $F_n(x,y)''>0$ follows from $\{y > x^q \leftrightarrow q > 2n: q \in \mathbf{Q}\}$. If we take $x,y,n$ as real variables these are all statements in the theory of real-closed fields. Then we apply several properties that theory: since this is true for infinitely many real $n$, it must be true for all sufficiently large $n$, it must follow from a finite subset of the hypotheses, and it must be provable. It should be possible to adapt @PeterMueller's proof along these lines. | |
| Jan 13, 2017 at 13:04 | comment | added | Pietro Majer | Sure, for $F(x,y)$, but I was talking about the composition, $F_n(x):=F(x,y(x))$, with $y(x):=x^{2n}$ as you said. So $F_n''(x)=D^2F[1,\dot y][1,\dot y]\ {\mathbf +}\ \partial_yF \ddot y$. | |
| Jan 13, 2017 at 12:11 | comment | added | Dima Pasechnik | A twice diff. function is convex on an open set $U$ iff its Hessian is positive semidefinite on $U$. I don't see why 1st partials matter at all.. | |
| Jan 13, 2017 at 10:56 | comment | added | Pietro Majer | And about $\partial_y F$? Don't we need it to be positive (for $0<y<x<1$) to conclude that $F(x,x^{2n})$ is convex? | |
| Jan 13, 2017 at 10:31 | history | answered | Dima Pasechnik | CC BY-SA 3.0 |