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EDIT: Let $\{t_n\}$ be a sequence of reals s.t. $t_n \rightarrow t_0$ with $t_n \ne t_0$.

Let $\{g_{t_n}\}$ be a sequence of integrable functions such that for $n$ large enough, $g_{t_n} \le g$ for some integrable $g$.

Then for a fixed function $g_{t_0}$, consider now the derived sequence of functions:

$\left\{ \frac{(g_{t_n} - g_{t_0})}{t_n - t_0} \right\} = \{d_n\}$


How do I show that $\{d_n\}$ is also dominated for $n$ large enough by some other function (no doubt derived in some way from $h$)?

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up vote 1 down vote accepted

If $t_0$ is a limit point of $(t_n)$, the result is doubtful. If $|t_n-t_0|\geqslant\varepsilon$ for every $n$, take $2h/\varepsilon$.

For an example of the first assertion, start from some integrable function $g$ such that $g'$ is not integrable, say $g(x)=\sin(x^2)/(1+x^2)$, and define $t_n=1/n$, $t_0=0$, $g_{t}(\cdot)=g(t+\cdot)$.

Then every $g_{t_n}$ is a shift of $g$ by at most $1$, $|g|\leqslant f$ with $f(x)=1/(1+x^2)$, and an easy computation shows that $|g_{t}|\leqslant3f$ for every $|t|\leqslant1$, hence $(g_{t_n})$ is dominated by the integrable function $h=3f$. But $d_n\to g'$ when $n\to+\infty$ hence $(d_n)$ is not dominated by an integrable function although $(g_{t_n})$ is.

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$t_0$ is indeed a limit point of $(t_n)$ (original post now edited) – user1770201 Oct 28 '12 at 10:22
Please DO NOT DO THAT. You now modified substantially your post three times, each time asking a different question. People answering your post are not supposed to follow your wandering and to adapt their answers to your latest whims. Please revert to a previous version and ask another question in another post, if you wish. (Besides, the current version of the question is absurd.) – Did Oct 28 '12 at 13:01

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