Hmmm Arzela-Ascoli, maybe?

[AxiomOfChoice]

Suppose you know the following about g: \mathbb R \to \mathbb R:

1. g\in C^1;
2. g\in C^2, expect at finitely many points \{x_1,\ldots,x_n\}, and |g''(x)| \leq M (except at those points).

How can you show that there exists a sequence f_k with the following properties?

1. f_k \to g uniformly;
2. f'_k \to g' uniformly;
3. f_k \in C^2, |f''_k(x)| \leq M, and f''_k \to g'' outside \{x_1,\ldots,x_n\}.

Arzela-Ascoli is pretty much the only theorem I know of that talks about sequences of functions with uniformly bounded derivatives, but I can't for the life of me see how you could use it to construct the sequence f_k.

 

[AxiomOfChoice]

Hmmm...well, what if instead of looking at \mathbb R, we restrict our attention to some compact subset of \mathbb R? Does that help at all?

 

[micromass]

Take f_k=g for each k. Or do you want f_k to be C^2 at each point?? In that case, first approximate g^{\prime\prime} by continuous functions. Then take integrals.

 

[AxiomOfChoice]

Take f_k=g for each k. Or do you want f_k to be C^2 at each point?? In that case, first approximate g^{\prime\prime} by continuous functions. Then take integrals.

micromass, thanks for your response.

Since g'' is continuous (except on a set of measure zero) on the compact interval [0,t], it is integrable there and can therefore be written as the pointwise limit of a sequence of continuous functions (with compact support) - call these functions f''_k. We know these functions are integrable, too, and that f'_k(s) = \int_0^s f''_k(q)dq for s\in [0,t]. We also have
<br /> |f&#039;_k(s)| = \left| \int_0^s f&#039;&#039;_k(q)dq \right| \leq \int_0^s |f&#039;&#039;_k(q)|dq \leq Ms \leq Mt,<br />
such that the sequence \{f&#039;_k\} is uniformly bounded on [0,t]. But how are we supposed to show that f&#039;_k \to g&#039; uniformly? I mean, do we even know \int_0^s g&#039;&#039;(q)dq = g&#039;(s)?

 

[micromass]

micromass, thanks for your response.

Since g&#039;&#039; is continuous (except on a set of measure zero) on the compact interval [0,t], it is integrable there and can therefore be written as the pointwise limit of a sequence of continuous functions (with compact support) - call these functions f&#039;&#039;_k. We know these functions are integrable, too, and that f&#039;_k(s) = \int_0^s f&#039;&#039;_k(q)dq for s\in [0,t]. We also have
<br /> |f&#039;_k(s)| = \left| \int_0^s f&#039;&#039;_k(q)dq \right| \leq \int_0^s |f&#039;&#039;_k(q)|dq \leq Ms \leq Mt,<br />
such that the sequence \{f&#039;_k\} is uniformly bounded on [0,t]. But how are we supposed to show that f&#039;_k \to g&#039; uniformly?

So you must prove that \int_0^s f^{\prime\prime}_k\rightarrow \int_0^s g^{\prime\prime} uniformly. This is easily proved from the definition of uniform continuity. Use that

|\int_0^s{f^{\prime\prime}_k} - \int_0^{s-h}{g^{\prime\prime}}| \leq \int_{s-h}^s{|f^{\prime\prime}_k - g^{\prime\prime}|} \leq h \sup_{x\in [s-h,h]}{ |f^{\prime\prime}_k(x)-g^{\prime\prime}(x)|}

I mean, do we even know \int_0^s g&#039;&#039;(q)dq = g&#039;(s)?

Fundamental theorem of calculus? (if you add a -g'(0) there)

 

[AxiomOfChoice]

Fundamental theorem of calculus? (if you add a -g'(0) there)

But g&#039;&#039; is not continuous at finitely many points in the interval, so we can't apply the (traditional) FTC. Maybe we can apply Lebesgue's FTC, but how do we know g&#039;&#039; is absolutely continuous?

 

[micromass]

But g&#039;&#039; is not continuous at finitely many points in the interval, so we can't apply the (traditional) FTC. Maybe we can apply Lebesgue's FTC, but how do we know g&#039;&#039; is absolutely continuous?

Continuity if g^{\prime\prime} isn't necessary.

It can be proven that if f:[a,b]\rightarrow \mathbb{R} has a primitive F on [a,b], then

\int_a^b{f}=F(a)-F(b)

Continuousness of f isn't important. This is a slight generalization of the usual fundamental theorem of calculus. A proof can be found in Bartle "a modern theory of integration" Theorem 4.5.