Timeline for Inequalities between $M_k=\sup \limits_{\mathbb{R}}|f^{(k)}(x)|$
Current License: CC BY-SA 4.0
20 events
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| Apr 11, 2021 at 23:49 | comment | added | RFZ | @GiorgioMetafune, Indeed. I completely forgot that we are given that $M_0<\infty$. Thanks a lot for your help! I solved this question. | |
| Apr 11, 2021 at 7:49 | comment | added | Giorgio Metafune | You have that $\sup |f(x)| \leq C$, then apply the estimates to $f/C$. | |
| Apr 11, 2021 at 0:23 | comment | added | RFZ | @GiorgioMetafune, thanks a lot for your reply! I think that I need to clarify one more question: You said that we can apply 1b) to estimate $m_k$, right? But in order to use 1b) we need to have $\sup \limits_{(-1,1)}|f(x)|\leq 1$. But in our case it may not be true. Could you give more details, please? | |
| Apr 4, 2021 at 7:22 | comment | added | Giorgio Metafune | By assumption you have $M_0(I), M_p(I) \le C$ for every unit interval $I$ and, by the argument above applied to each $I$, $M_k(I) \le C_k$, for every $I$. | |
| Apr 4, 2021 at 1:56 | comment | added | RFZ | @GiorgioMetafune, yesterday I was thinking on this problem. More precisely, on your hint. Let me ask you question please. You said that "This gives the result in all unit intervals hence on the line but with a worse bound". Can you clarify this sentence, please? If the function is bounded on all unit intervals how it follows that it is bounded on $\mathbb{R}$? For example, this is not true for $f(x)=x$. | |
| Mar 30, 2021 at 17:24 | comment | added | Giorgio Metafune | Use 1b) and call $m_k, M_k$ the minimum and the maximum of $|f^k|$ in the whole $[0,1]$. The $m_k$ are estimated in point 1b) and from the mean value theorem you get $M_k \leq m_k+M_{k+1}$. If you assume that $M_p \leq C$, then inductively all $M_k$ are estimated. This gives the result in all unit intervals hence on the line but with a worse bound. At this point you know that all derivatives are bounded and run the induction to improve the bounds. | |
| Mar 30, 2021 at 16:44 | comment | added | RFZ | @GiorgioMetafune, I did not get the idea of what you said. But of you have some free time that would be great to see more detailed solution of what you suggested. I'll highly appreciate your help! | |
| Mar 30, 2021 at 16:40 | comment | added | Giorgio Metafune | You are write, I did not know and I did the simplest thing using more tools. However also the induction steps are tricky. You can show that a bound for the function and the highest order derivative also gives a bound for intermediate derivatives on an interval, with a constant depending on the interval. Then split the whole line into a sequence of unit intervals and get the global bound, but worse of what you want. Then run the induction as above | |
| Mar 30, 2021 at 16:33 | comment | added | RFZ | Wow. I supposed that it problem can be solved much easier since this is just a problem for first year undergrad students. | |
| Mar 30, 2021 at 16:31 | comment | added | Giorgio Metafune | Yes, sure. The map $$a=(a_0, \dots, a_n) \to \|\sum_{k=0}^n a_k h^k\|_\infty$$ is continuous and strictly positive on the unit sphere $\|a\|=1$ of $R^n$ (by the identity principle for polynomials) and hence has a positive minimum $\delta>0$. By homogenuoity this gives the inequality above with $C=1/\delta$. | |
| Mar 30, 2021 at 16:27 | comment | added | RFZ | @GiorgioMetafune, I guess your last comment is incomplete :) | |
| Mar 30, 2021 at 16:17 | comment | added | RFZ | @GiorgioMetafune, please sorry I am still confused. How did you get the above inequality? Could you give more details please? | |
| Mar 30, 2021 at 14:58 | comment | added | Giorgio Metafune | I am just using that $$|a_k| \le C\|\sum_{k=0}^n a_k h^k\|_\infty$$ (sup norm on some interval). However there are other arguments: one could approximate with convolutions and using the apriori bound or localizing to intervals as the exercise suggests. | |
| Mar 30, 2021 at 14:29 | comment | added | RFZ | @GiorgioMetafune, this is just a problem from real analysis for undergraduates. I'll prefer if we will not use heavy machinery. | |
| Mar 30, 2021 at 14:27 | comment | added | RFZ | @GiorgioMetafune, I am confused, so please let me ask you some questions. I will try to expand your last comment (please correct me if I'm wrong): for any $x\in \mathbb{R}$ and $0<|h|\leq 1$ by Taylor's formula we have: $f(x+h)=f(x)+f'(x)h+\dots+\dfrac{f^{p-1}(x)}{(p-1)!}h^{p-1}+\dfrac{f^{(p)}(\xi)}{p!}h^p$ for some $\xi$ in between $x$ and $x+h$. Denote $P(h):=f(x+h)-f(x)-\dfrac{f^{(p)}(\xi)}{p!}h^p=\sum \limits_{k=1}^{p-1}\dfrac{f^{(k)}(x)}{k!}h^k$ and the LHS can be bounded in the following way: $|P(h)|\leq |f(x+h)|+|f(x)|+|f^{(p)}(\xi)|\leq 2C.$ How it follows that $|f^{(k)}(x)|\leq K$? | |
| Mar 30, 2021 at 8:45 | comment | added | Giorgio Metafune | Yes, true, but this follows from this argument. Call $C=\|f\|_\infty+\|f^p\|_\infty$; by Taylor's formula the polynomial $$P(h)=\sum_{k=1}^{p-1} \frac{f^k(x)}{k!}h^k$$ satisfies $|P(h)| \le 2C(1+|h|^p) \le 4C$ if $|h| \leq 1$. Since all norms on a finite dimensional space are equivalent, then $|f^k(x)| \le K$, with $K$ independent of $x$. | |
| Mar 30, 2021 at 1:17 | comment | added | RFZ | @GiorgioMetafune, I think that I got how to prove it but I guess there is one subtle moment. More precisely, in this proof we are implicitly using the fact that $M_k$ are are finite. But we do not know this in advance. | |
| Mar 29, 2021 at 7:14 | comment | added | Giorgio Metafune | With $C=\sqrt 2$, show first that $M_k \le C^k M_0^{\frac{1}{k+1}} M_{k+1}^{\frac{k}{k+1}}$ by induction on $k$. | |
| Mar 29, 2021 at 3:28 | review | Close votes | |||
| Apr 10, 2021 at 3:02 | |||||
| Mar 29, 2021 at 1:22 | history | asked | RFZ | CC BY-SA 4.0 |