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
$$|f'_k(s)| = \left| \int_0^s f''_k(q)dq \right| \leq \int_0^s |f''_k(q)|dq \leq Ms \leq Mt,$$
such that the sequence $\{f'_k\}$ is uniformly bounded on $[0,t]$. But how are we supposed to show that $f'_k \to g'$ uniformly? I mean, do we even know $\int_0^s g''(q)dq = g'(s)$?

 

[micromass]

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
$$|f'_k(s)| = \left| \int_0^s f''_k(q)dq \right| \leq \int_0^s |f''_k(q)|dq \leq Ms \leq Mt,$$
such that the sequence $\{f'_k\}$ is uniformly bounded on $[0,t]$. But how are we supposed to show that $f'_k \to g'$ 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 $\int_0^s g''(q)dq = g'(s)$?

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

 

[AxiomOfChoice]

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

But $g''$ 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 $g''$ is absolutely continuous?

 

[micromass]

But $g''$ 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 $g''$ 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.