Timeline for Proving the existence of a continuous function that satisfy a certain property from a finite version of this property
Current License: CC BY-SA 3.0
8 events
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| Nov 9, 2017 at 21:04 | comment | added | Eilon | Thanks. In my original question I had $\mathbb{R}^n_+$ instead of $\mathbb{R}^n$ throughout, and I thought that changing this to $\mathbb{R}^n$ would not matter. In fact, it does not, since my son came up with a counterexample to the case where we require $M$ to be a subset of the nonnegative orthant: take $M = (t,x,(t-\frac{1}{2})^2)$, where $t,x \in [0,1]$ and $\frac{x}{3} \leq t \leq \tfrac{2+x}{3}$. | |
| Nov 9, 2017 at 4:25 | comment | added | fedja | That is obviously false as written now: take $n=1$ and $M=\{(y-x+0.5)^2\le 0.01\}$. Then at every particular point you have an interval of length $0.2$ of admissible values, so you can choose a value of size at least $0.1$ and extend linearly between the points but your $f$ has to change sign on $[0,1]$, so the intermediate value theorem is a killer. I surmise you meant something else. | |
| Nov 8, 2017 at 14:38 | comment | added | Dirk | Oh, I missed that $M\subseteq [0,1]\times \mathbb{R}^n$ (somehow I assumed that $M\subseteq \mathbb{R}^n$…) | |
| Nov 8, 2017 at 12:01 | history | edited | Eilon | CC BY-SA 3.0 |
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| Nov 8, 2017 at 12:00 | comment | added | Eilon | Dear Dirk, I would not mind having a constant function, as long as its graph is a subset of M (and the constant is above $\delta_0$). | |
| Nov 8, 2017 at 10:05 | comment | added | Dirk | A constant function $x$ is not what you want, right? | |
| Nov 8, 2017 at 9:44 | history | edited | Eilon | CC BY-SA 3.0 |
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| Nov 8, 2017 at 9:09 | history | asked | Eilon | CC BY-SA 3.0 |