Given a bounded and uniformly continuous function $u:\mathbb{R}^n \to \mathbb{R}$, is there a family of equi-bounded $C^\infty$ functions $\{u_\epsilon\}$ with equi-bounded partial derivatives that converges to $u$ uniformly?
Bounded $C^\infty$ functions $\{u_\epsilon\}$ with bounded derivatives s.t. $\to$ uniformly to bounded and unif. cont. $u:\mathbb{R}^n \to \mathbb{R}$
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real-analysis
functional-analysis
1 Answers
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If $u$ is not bounded, it is impossible for a sequence of bounded functions to converge uniformly to $u$.
EDIT: For the revised question, the answer is still no. Consider e.g. $u(x) = \cos(x^2)$, which is bounded and has $|u(\sqrt{n\pi}) - u(\sqrt{(n+1)\pi})| = 2$. Note that $\sqrt{(n+1)\pi} - \sqrt{n\pi} \to 0$ as $n \to \infty$. If $|v'| \le B$, then for sufficiently large $n$ we have $|v(\sqrt{n\pi}) - v(\sqrt{(n+1)\pi})| \le B (\sqrt{(n+1)\pi} - \sqrt{n\pi}) < 1$, so one of $|v(\sqrt{n \pi} - u(\sqrt{n\pi})|$ and $|v(\sqrt{(n+1)\pi}) - u(\sqrt{(n+1)\pi})|$ is greater than $1/2$.
EDIT: With the added assumption of uniform continuity, the answer is yes. Take convolutions of $u$ with a sequence of functions $v_n \ge 0$ which are $C^\infty$, supported in $\{x: \|x\| < 1/n\}$, and have $\int v_n = 1$.
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0That's right, of course. I edited the question. – 2017-01-12
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0Thanks. And if we assume $u$ bounded and *uniformly continuous*? – 2017-01-12
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0Thank you. Could you add some details to your answer? In particular: what is an example of a family of such bump functions $\nu_\epsilon$? How do we prove that the family $u_\epsilon = \nu_\epsilon \ast u$ is $C^\infty(\mathbb{R}^n)$, bounded and has bounded partial derivatives and converges to $u$ uniformly? – 2017-01-13
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0Also, is it possible to have $u_\epsilon$ equibounded and with derivatives equibounded? – 2017-01-13