Let $f \in C^k_b([0,\infty); \mathbb{R})$. We want to extend $f$ to the space $\mathbb{R}$ without loosing regularity and boundedness (thanks zhw. for this observation), i.e. we want some $F \in C^k_b(\mathbb{R}; \mathbb{R})$ with $F|_{[0,\infty)} = f$. First idea to come into ones mind is to simply reflect, i.e. to set $$ F(x) = \begin{cases} f(x) &\quad x \geq 0 \\ f(-x) &\quad x < 0\end{cases} $$

This gives us $F(x) \in C(\mathbb{R}; \mathbb{R})$ but higher derivatives needn't be continuous. The idea of higher order reflections is to write $$ F(x) = \sum_{j = 1}^{k + 1} \lambda_j \; f(-jx), \quad (x < 0) $$ and determine the linear factors $\lambda_j$ such that $D^mF$ is continuous at $x= 0$ for all $m \in \{0,\dots,k\}$. This gives the following set of equations \begin{align} f(0) &= F(0) = \sum_{j = 1}^{k + 1} \lambda_j \; f(0) \\ f'(0) &= F'(0) = \sum_{j = 1}^{k + 1} (-j) \lambda_j \; f'(0) \\ &\dots \\ f^k(0) &= F^k(0) = \sum_{j = 1}^{k + 1} (-j)^k \lambda_j \; f^k(0) \end{align}

How would you motivate this procedure? The only references I found so far were this blog post and this other post which are a good start but not completely satisfying. In both posts it is said that to preserve the $C^k$ smoothness, the extension process must preserve polynomials of x of degrees up to $k$. Could someone elaborate on this fact?

Bottom line so far: An extension operator which is a convex combination of "scaling" operators", i.e. some operator $$ E(u) := \sum_{j = 1}^{k + 1} \lambda_j \; \delta^{-j}(u), $$ where $\delta^{-j}(u)(x) := u(-jx)$ seems to be an apt candidate for an extension operator that maintains differentiability. Is there some intuition to this? Why would one think this is a good idea, other than making a direct calculation.

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    $\begingroup$ Why not extend by defining $$f(x) = \sum_{n=0}^{k}\frac{ f^{(n)}(0) x^n}{n!}$$ to the left of $0$? $\endgroup$
    – zhw.
    Jul 26, 2017 at 16:10
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    $\begingroup$ @zhw. Good point. However, this will not give a bounded extension for bounded $f$. I will edit my post. $\endgroup$
    – el_tenedor
    Jul 26, 2017 at 16:14

1 Answer 1


the extension process must preserve polynomials of $x$ of degrees up to $k$

Yes, and here is why: suppose $f$ is actually a polynomial of degree $\le k$ on the interval $[0,1]$, continued in some smooth and bounded fashion to $[0,\infty)$. Then the reflection

$$F(x) = \sum_{j = 1}^{k + 1} \lambda_j \; f(-jx), \quad (x < 0)$$

is also a polynomial of degree $\le k$ on the interval $(-1/(k+1), 0)$.

Key point: a piecewise function where both pieces are polynomials of degree $\le k$ is not $C^k$ smooth unless both polynomials are the same.

So, we must have

$$x^m = \sum_{j = 1}^{k + 1} \lambda_j \; (-jx)^m, \quad 0\le m\le k$$

This linear system is solvable because the Vandermonde matrix is inverible.

It remains to prove that the resulting extension actually works for all $C^k$ functions. This can be done by differentiating $F$ repeatedly and observing what we get as $x\to 0-$.

Alternatively, a computation-free proof. We know that if $f$ is extended by some linear combination of reflections, and the $0$-degree Taylor polynomials match at $0$, then the extension is continuous. Proceed by induction: $F'$ is a linear combination of reflections of $f'$, and their Taylor polynomials at $0$ match up to order $(k-1)$. This shows that the one-sided limits of derivatives agree. Finally, recall a useful fact from real analysis: if $g$ is continuous at $0$ and $\lim_{x\to 0}g'(x)$ exists, then $g'(0)=\lim_{x\to 0}g'(x)$.

Why would one think this is a good idea

What other ideas do we have? (Except just using a polynomial like in zhw.'s comment). Using $f(-x)$ is the standard thing to do, but one needs more to achieve smoothness, and what other ingredients do we have except $af(-bx)$?


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