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.

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 $|f'(x) - f'(y)| \leq C|x - y| \hspace{1 pc}\forall x, y \in [a, b]$.

share|improve this question
    
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
show 6 more comments

1 Answer

up vote 1 down vote accepted

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 $f^{'}_{n_k} \to f^{'}.$

share|improve this answer
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.