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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Hi could you help me with the following:

If I have a function $g \in C^2(0,R)$ with $|g''(x)| \le M$, i.e. its second derivative is bounded except a finite number of points where at those irregularity points none of the assumptions mentioned above holds can you give me a sequence of functions $f_n$ with $f_n^\prime \longrightarrow g'$ and $f_n \longrightarrow g$ uniformly and $|f_n^{\prime\prime}(x)| \le M$?

I think about polynomials but can not justify that they have all the requirements??

Thanks a lot!!

share|cite|improve this question
What kind of functions? $f_n = g$ for all $n$ is a trivial example. – Lukas Geyer Oct 8 '12 at 0:35
I am sorry i should have added g∈C2(0,R) except finite points x_1, x_2, .. ,x_n. The question is how to find a sequence f_n which are completely nice with those properties? Thank you – Salih Ucan Oct 8 '12 at 1:26
You can (and should) edit your post to add this information. Also, what kind of regularity does the function at these points have? Is it still continuous and differentiable? Is the derivative continuous there? – Lukas Geyer Oct 8 '12 at 5:31

Think of the following functions: $$f_n(x) = \left\{ \begin{matrix} g_n(x), \mbox{if } x \in I, \\ l_k, \mbox{ if } x \notin I. \end{matrix} \right.$$ Where $I \subset \left[0,2 \right]$ denotes the subset in which the conditions hold, and $l_k = \lim_{x \to x_k}g_n(x)$, being $x_k$ the points outside $I$. Since the $g_n$ are continuous almost for every point, their limits on those irregular points still exist, so you can use them to make the $f_n$ continuous.

Now, the $f_n$ are continuous in a compact set, so they are bounded and they are obviously $\mathcal{C}^2$. Plus, $\left| f''_n (x) \right| \leq \left| g''(x) \right| \leq M$, so I guess it will do the trick.

I hope this helps.

share|cite|improve this answer

Your Answer


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.