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Are the following statements true? An answer with only "Yes" or "No" for each number is acceptable for me.

  1. Let $f: U \to \mathbb R$ be continuous with $U$ nonempty and open in $\mathbb R^n$. Then $\tilde{f} = f1_U+0 1_{U^c}$, the extension by zero of the function $f$, is not necessarily continuous. However, there exists $\rho$, a bump function at a point $q \in U$, such that $\hat{f} = f\rho1_U+0 1_{U^c}$ is continuous and agrees with $f$ on some neighborhood $W$ of $q$ but not necessarily on the whole of $U$.

    • This is supposedly an analogue of (2) and (3) here for continuous in the case that $M=\mathbb R^n$

  2. (Analogue of (2) and (3) for $C^k$, $M=\mathbb R^n$) Let $f: U \to \mathbb R$ be $C^k$ with $U$ nonempty and open in $\mathbb R^n$. Then $\tilde{f}$ is not necessarily $C^k$. However, there exists $\rho$, a bump function at a point $q \in U$, such that $\hat{f} = f\rho1_U+0 1_{U^c}$ is $C^k$ and agrees with $f$ on some neighborhood $W$ of $q$ but not necessarily on the whole of $U$.

    • The $k=\infty$ case is (2) and (3) here, and the $k=0$ case is (16). (17) asks about $0 < k < \infty$.

Context:

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    Didn't you mean 'but not necessarily on the whole U'? If so then product of $C^{k}$ with smooth function is at least $C^{k}$. To find correct bump take smooth approximation of $f$ and use fact that those are dense – Radost May 09 '19 at 00:03
  • @Radost Thanks! Are the statements wrong? I just changed $C^{\infty}$ to $C^0$ and to $C^k$, but I used the same argument, so I guess you mean that the arguments are wrong even if the extensions exist –  May 09 '19 at 00:10
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    Statements are true. Taking smooth indicator of some closed set contained within $U$ and containing $q$ the product is at least as smooth as the original function. Existence of such functions is discussed on wikipedia for example. – Radost May 09 '19 at 00:26
  • @Radost Thanks! –  May 09 '19 at 01:12

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