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So this thought popped into my head while attempting to solve a grad-level real analysis assignment. Though this question is probably basic, I am finding it hard to justify a claim I am making (I am not a math major).

Firstly, the question I am trying to tackle basically states the following: $\{f_n, n = 1, ...\}$ are real valued and twice differentiable. They also converge uniformly to an arbitrary function, $f$, and the second derivatives, $f_n''$, are uniformly bounded. I essentially have to show that the first derivative, $f_n'$, converges uniformly to $f'$. I suppose this is a standard problem (precisely, this is the Lipschitz condition).

I was thinking in the lines of proof by contradiction. At first, I assumed $f_n'$ doesn't converge to $f'$. So, roughly speaking, there will be some $x^*$ where the difference between $f_n(x)$ and $f_m(x)$ is greater than some $\epsilon > 0$ $\forall n,m > n_0$ for some $n_0 \in \mathbb{N}$. Then, I wished to consider a $\delta$-neighborhood of $x^*$. Then, for all $x$ in this neighborhood, without loss of generality, if we assume that $f_n'$ converges to $f'$ pointwise, then I intuitively think that the second derivative should become arbitrarily large at that point. Because, for all $x$ in this neighborhood, $| f_n'(x) - f_m'(x)| < \epsilon$ (for sufficiently large $n, m$) while but at $x^*$ it is greater than the same $\epsilon$ (where I pick $\epsilon$ as in the definition of $x^*$).

I am pretty sure that to make the above argument rigorous, I need to use the fact that the second derivative is uniformly bounded. I also have a hunch that if I can argue that since the second derivative is uniformly bounded, then for any chord (that is line segment joining two points of a curve of a function in $\mathbb{R}$, just my "term" for this) then I can pick one of the two points to be $x^*$ and the other point to be in the $\delta$-neighborhood and then somehow use this property to write in mathematical terms the argument I presented above.

I want to know if my argument is sound and if I can indeed claim the above property via uniform boundedness and if indeed the above technique can prove my argument mathematically.

A clarification of the original question: The question needs me to show that $f_n' \rightarrow f'$ uniformly and that $\exists C > 0$ such that $|f'(x) - f'(y)| \leq C|x - y| \hspace{1 pc}\forall x, y \in [a, b]$.

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You've assumed $f_n'$ does not converge to $f'$, but for contradiction, you need to assume they don't converge uniformly. I think you can prove this in the positive direction. Hint: the fact that $f_n''$ are uniformly bounded means that the $f_n'$s are uniformly Lipschitz. That is, $\exists M : \forall x,y,n, |f_n'(x) - f_n'(y)| < M|x - y|$. –  BaronVT Sep 8 '12 at 0:12
Hi, the fact that you have used is exactly what I need to prove! I can't use the same thing to prove something! Moreover, I realized that the converse is not exactly what I did, but nor is the converse what you stated. Specifically, a sequence of functions can converge pointwise and still not be uniformly convergent. And if a sequence of functions don't converge pointwise (what I assumed) they will NOT converge uniformly. Assuming they don't converge uniformly seems to implicitly assume at least convergence, which itself is something I need to prove... I may be doing something wrong however. –  Abhijit Sep 8 '12 at 0:23
On what domain are the functions $f_n, f$ defined? –  Nate Eldredge Sep 8 '12 at 0:27
Think about your logic: you are trying to say "assume they do not converge pointwise. Then ... contradiction." This would only prove that they do converge pointwise; it would not prove uniform convergence. –  Nate Eldredge Sep 8 '12 at 0:29
The domain of $f_n$ and $f$ is $\[a, b\] \subseteq \mathbb{R}$ . I haven't noticed this myself! Can this help? I agree with you Nate. This was going to be the first step of my proof. I hoped to show uniform convergence in a later step. –  Abhijit Sep 8 '12 at 0:29
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1 Answer

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Why don't you first try and show that a subsequence of $f^{'}_n$ converges uniformly to $f^{'}$? I think this is an application of a well known theorem called the Arzela Ascoli theorem. The Arzela-Ascoli theorem states that if we have a sequence $\{f_n\} \in C[a,b]$ such that $\{f_n\}$ is uniformly bounded and equicontinuous, then there exists a subsequence$\{f_{n_k}\} \to f \in C[a,b]$ uniformly. We are now going to show the uniform boundedness and equicontinuity of $f^{'}_n$. This will tell us that there exists $f^{'}_{n_k}$ that converges uniformly to $g \in C[a,b]$. We can then show that $g = f^{'}$

We know there exists $\gamma \in [a,b]$ such that $$|f^{'}_n(x) - f^{'}_n(y)| = |f^{''}_n(\gamma)||x-y| \leq M|x-y|$$ by Taylor's theorem and the fact that $f^{''}_n$ is uniformly bounded. Thus, $f^{'}_n$ is equicontinuous.

To show uniform boundedness of $f^{'}_n$ we do the following: $$|f^{'}_n(x)| = |f^{'}_n(x) - f^{'}_n(0) + f^{'}_n(0)| = |f^{''}_n(\gamma)x + f^{''}_n(0)| \leq M(1+|x|).$$ The above follows by Taylor's theorem and triangle inequality (and the fact that $f^{''}_n$ is uniformly bounded.) Thus, by Arzela Ascoli, there exists a subsequence $\{f_{n_k}\} \to g$, $g \in C[a, b]$

For the rest note that $$f_{n_k}(x) - f_{n_k}(a) = \int_{a}^{x}{f^{'}_{n_k}dt}$$ Since the $f^{'}_{n_k}$ converge uniformly, so does the integral. Thus, $f(x) - f(0) = \int_{a}^{x}{g(t)dt}$. By the Fundamental Theorem of Calculus, $g = f'$ so $f^{'}_{n_k} \to f^{'}.$

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