Constructing a sequence of function with bounded derivative

Question

Let $f:\mathbb R\mapsto\mathbb R$ be a smooth function and analytic at $x=0$. I wish to find a sequence of functions $\{f_n\}$ such that $\{f_n(x)\}$ is convergent to $f(x)$ for all $x$ and $f'''_n$ is bounded. I know, we could take the Taylor polynomial if there was no condition $f'''_n$. However, this boundedness rises the difficulty of my problem. Any suggestion?

Answer 1

First, my version of your question: Take $f$ a function, say continuous on $\mathbb{R}$, analytic at $x=0$. You want to show that there exists a sequence $f_n$ of functions, say $\in C^3(\mathbb{R})$, such that for all fixed $x\in \mathbb{R}$, we have $f_n(x)\to f(x)$, and that there exists $M>0$, such that for all $n$ and $x\in \mathbb{R}$, we have $|f_n^{(3)}(x)|\leq M$. (See your answer to ChristopherA.Wong comment)

If this is your question, I think that additionnal hypothesis are needed.

To see why, let $\alpha_k$ be constants, not all zero, such that $\alpha_0+\alpha_1+\alpha_2+\alpha_3=0$, and $\alpha_1+2\alpha_2+3\alpha_3=\alpha_1+2^2\alpha_2+3^2\alpha_3=0$.

We have for $t=1,2,3$ that there exists $c_t\in \mathbb{R}$ (depending on $h,x,n..$) such that: $$f_n(x+th)=f_n(x)+thf_n^{\prime}(x)+\frac{t^2h^2}{2}f_n^{\prime\prime}(x)+\frac{t^3h^3}{6}f_n^{(3)}(c_t)$$

We get: $$\alpha_0 f_n(x)+\alpha_1 f_n(x+h)+\alpha_2 f_n(x+2h)+\alpha_3f(x+3h)=(\alpha_1f_n^{(3)}(c_1)+2^3f_n^{(3)}(c_2)+3^3\alpha_3f_n^{(3)}(c_3))\frac{h^3}{6}$$ Now, by the hypothesis, there exists $M_1$ independant of $n,x,h$, such that for all $n$, $x$ and $h$: $$|\alpha_0 f_n(x)+\alpha_1 f_n(x+h)+\alpha_2 f_n(x+2h)+\alpha_3f_n(x+3h)|\leq M_1|h|^3$$

and if $n\to +\infty$ we get: $$|\alpha_0 f(x)+\alpha_1 f(x+h)+\alpha_2 f(x+2h)+\alpha_3f(x+3h)|\leq M_1|h|^3$$ for all $x$ and $h$. You can see that this condition is not satisfied by $f(x)=\exp(x)$ for example.