Approximation of piecwise linear function where $f_n(x)=0$

Question

Given $I:=(-\infty, 2)$, I am looking at the function $f:I \rightarrow \mathbb{R}$, $$ f(x) := x \chi_{[0, \infty)}(x). $$ I am well aware that this is a function in $W^{1, 1}(I)$, so there must be some sequence $$(f_n)_{n \in \mathbb{N}} \subseteq C^3((-\infty,2]) \cap W^{1, 1}(I)$$ (even $C^\infty$, but $C^3$ is enough for my purposes) such that $$ \lVert f_n - f \rVert_{W^{1, 1}(I)} \rightarrow 0 $$ as $n \rightarrow 0$.

I am wondering whether one can ensure that $f_n(x) = 0$ for all $x \in (-\infty, -\varepsilon)$ where $\varepsilon > 0$ is small and independent of $n$. I tried to find approximating functions by spline interpolation, but that does not seem to work because of the discontinuity in the derivative. I also tried to only approximate $\chi_{[0, \infty)}$ in $C_0^\infty$, but that leads me to nowhere - I do not know whether the derivative converges.

Am I missing something obvious?

Answer 1

There are a few ways to resolve this, but generally your goal will be to define a sequence of functions $f_n \in C^\infty$ (or $C^3$, or whatever) such that $f_n = f$ outside $(-1/n, 1/n)$ and there is some $M$ independent of $n$ such that $|f_n| + |f_n'| \leq M$ inside $[-1/n, 1/n]$. These aren't necessary conditions, but they are sufficient and they're general enough to guide your choices.

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Using polynomial interpolation to create a sequence in $C^3$:

We try polynomial [Hermite] interpolation at the endpoints of so that $$f_n(x) = \begin{cases}P_n(x) = \sum_{k=0}^7 a_kx^k,\quad & x\in (-1/n, 1/n) \\ f(x), & \text{otherwise}\end{cases}$$ such that $$P_n(1/n) = 1/n, \qquad P_n'(1/n) = 1, \qquad P_n''(1/n) = P_n'''(1/n) = 0\\ P_n(-1/n) = P_n'(-1/n)=P_n''(-1/n) = P_n'''(-1/n) = 0$$ which I find easier to solve by noting it satisfies the identity $P_n(x) = \frac1nP_1(nx)$. This allows us to note that the solution for $P_1$ gives us the solution for all $n$ and $$\sup_{-1/n \leq x \leq 1/n} |P_n(x)| = \frac1n\cdot \sup_{-1 \leq nx \leq 1} |P_1(nx)| \\ \sup_{-1/n \leq x \leq 1/n} |P_n'(x)| = \sup_{-1 \leq nx \leq 1} |P_1'(nx)|$$ so that on $[-1/n, 1/n]$, $$|f_n| + |f_n'| \leq \sup_{-1 \leq y \leq 1} |P_1(y)| + \sup_{-1 \leq y \leq 1} |P_1'(y)| := M$$ and therefore $$\begin{align*}\|f_n - f\| &= \|(f_n-f)\chi_{[-1/n, 1/n]}\| \\ &\leq \|f_n\chi_{[-1/n, 1/n]}\| + \|f\chi_{[-1/n, 1/n]}\| \\ &= \int_{-1/n}^{1/n} (|f_n(x)|+|f'_n(x)|)dx + \int_0^{1/n}(x+1)dx \\ &\leq \frac{2M}n + \frac1{2n^2} + \frac1n \to 0\end{align*}$$

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Using mollification to create a sequence in $C^\infty$:

To consider the same problem in $C^\infty$, it is probably best to consider the convolutions of $f$ with suitably chosen mollifiers, and for this case we should extend $f : \mathbb{R} \to \mathbb{R}$ by also defining $f(x) = x\chi_{[0,\infty)}(x)$ outside $I$ as well.

In particular, if we write $$\varphi(x) = \begin{cases}Ae^{-1/(1-x^2)}, \quad & |x| < 1 \\ 0, & \text{otherwise} \end{cases}$$ where $A>0$ is chosen so that $\int_\mathbb{R} \varphi = 1,$ then defining $\varphi_n(x) = n\varphi(nx)$ gives us a positive, symmetric mollifier in $C^\infty$ supported on $[-1/n,1/n]$. We may then define $f_n = f \ast \varphi_n$

We directly check (using the mollifier's symmetry and support along with the piecewise linearity of $f$), that $f_n=f$ outside $(-1/n, 1/n)$, and as an immediate consequence of the usual theory, $f_n \in C^\infty$. In addition, we may control $\int_{-1/n}^{1/n}(|f_n-f|+|f_n'-f'|)$ by noting the following:

• $|f| \leq 1/n$ and $|f'| \leq 1$ on $(-1/n, 1/n)$.
• Since $|f| \leq 2/n$ on $(-1/n,1/n)+\operatorname{supp}\varphi_n$, it follows that $|f_n| \leq 2/n$ on $(-1/n, 1/n)$.
• To control $|f_n'|$, we may go to the definition: $$\begin{align*}f_n'(x) &= (f\ast \varphi_n)'(x) \\ &= \frac{d}{dx}\left(\int_\mathbb{R} f(x-t)\varphi_n(t)\,dt\right) \\ &= \lim_{h\to 0} \int_\mathbb{R}\left(\frac{f(x-t+h)-f(x-t)}{h}\varphi_n(t)\right)\,dt\end{align*}$$ and $$\left|\frac{f(x-t+h)-f(x-t)}{h}\varphi_n(t)\right| \leq \varphi_n(t)$$ is majorized by an integrable function which is independent of $h$, so we may bring the limit inside to see $f_n'(x) = (f'\ast \varphi_n)(x).$ (This is probably a well-known theorem, but while I had remembered $f_n' = f\ast \varphi_n'$, I didn't recall general conditions giving the result I showed here.) Therefore, $$|f_n'| \leq \sup |f'| = 1.$$

Putting it all together, this shows, $$\begin{align*}\|f_n-f\| &\leq \int_{-1/n}^{1/n} (|f_n| + |f| + |f_n'| + |f'|) \\ &\leq \int_{-1/n}^{1/n} (\frac2n + \frac1n + 1 + 1) \\ &= \frac2n(\frac3n+2) \to 0\end{align*}$$