# Uniform Convergence Weierstrass

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!!

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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.