Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$\fbox{Hypothesis}$

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:

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

$\fbox{Problem}$

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 $h$)?

share|improve this question
add comment

1 Answer 1

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.

share|improve this answer
    
$t_0$ 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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.