Is there an example of a family of functions, index by $k$, that is in $C_b^k(\mathbb R)$ but not in $C_b^{k+1}(\mathbb R)$ for arbitrary $k$?
$C_b^k(\mathbb R)$ is the space of functions with continuous and bounded derivatives up to $k$.
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$\begingroup$ This should be a duplicate? But I did not find the previous one. $\endgroup$– GEdgarSep 5, 2018 at 10:45
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1$\begingroup$ Let $f(x) = 0$ for $x<0$ and $f(x) = \arctan(x)^{k-1}$ for $x>0$ to get an example with bounded derivatives. $\endgroup$– WintherSep 5, 2018 at 10:51
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2$\begingroup$ You may want to know that $C^k(\Bbb R)$ is not the space of functions with bounded derivatives up to $k$, but the space of functions with continuous derivatives up to $k$. $\endgroup$– user562983Sep 5, 2018 at 10:54
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$\begingroup$ Doesn't $\arctan$ have bounded derivatives for all $k$ (with the bound increasing with $k$)? $\endgroup$– TohikoSep 5, 2018 at 12:45
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3 Answers
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Define $f$ by $$ f(x) = \begin{cases} (1-x^2)^{k+1} & |x|<1\\0 & |x|\ge1 \end{cases} $$ Then all derivatives of $f$ up to order $k$ are continuous (they are zero at $|x|=1$). But the $k+1$-st derivative does not exist at $|x|=1$.
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Take a continuous function with compact support but not differentiable function $f(x)$ and integrate it $k$ times:
Let $g_1(x) = \int_{-\infty}^{x} f(x) dx$ and define: $g_i(x) = \int_{-\infty}^{x} g_{i-1}(x) dx$
Use : Lebesgue's Differentiation Theorem for Continuous Functions to prove that the derivatives exist.
$g_k(x)$ is the required function.
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$F (x)=|x|^{\lambda }, ~~k<\lambda<k+1$
