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I can't find the counterexample of function with this criteria:

  1. Function $f$ such that $f'$ is absolute continuous in $[a,b]$, $f'' \notin L^2[a,b]$ but $f''$ is bounded in $[a,b]$.

  2. Function $f$ such that $f'$ is absolute continuous in $[a,b]$, $f'' \in L^2[a,b]$ but $f''$ is not bounded in $[a,b]$.

  • 1
    If $f''$ is bounded, it must be in $L^2[a,b]$.2012-10-16

1 Answers 1

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For #1, we note that if $f''$ is bounded, then $f'' \in L^p[a,b]$ for all $p\geq 1$. Hence no example exists.

For #2, Consider the following example $f:[0,1] \to \mathbb{R}$. Let $f(x) = \int_0^x t^2 \sin \frac{1}{\sqrt{t^3}} dt$. Then $f'(x) = \begin{cases} x^2 \sin \frac{1}{\sqrt{x^3}}, & x>0 \\ 0, & x = 0 \end{cases}.$ Continuing gives $ f''(x) = \begin{cases} 2 x \sin \frac{1}{\sqrt{x^3}} - \frac{3}{2 \sqrt{x}}\cos \frac{1}{\sqrt{x^3}} , & x>0 \\ 0, & x = 0 \end{cases}.$

We note that $|f''(x)| \leq 2 x + \frac{3}{2 \sqrt{x}}$, and $\int_0^1 (2 x + \frac{3}{2 \sqrt{x}}) \, dx = 4$, so $f''$ is integrable. It follows from Rudin's "Real & Complex Analysis" Theorem 7.21 that $f'(x) = \int_0^x f''(t) \, dt$, hence $f'$ is absolutely continuous. We also note that $f''$ is unbounded, and that $f, f' \in L^p[0,1]$ for all $p\geq 1$. If we let $\phi:[a,b] \to \mathbb{R}$ be given by $\phi(x) = f(a+x (b-a))$, then $\phi$ satisfies all the required conditions.