Question

Why is it difficult to find an elementary function that generally and suitably satisfies the below piecewise function, without reference to conditions, given continuous elementary functions \(f\) and \(g\) such that \(f(a)=g(a)\)?

\[ h:\mathbb{R}\to\mathbb{R}, h(x)=\begin{cases} f(x) & x\geq a \\ g(x) & x\leq a \end{cases}\]

Thoughts

Formulation?

I mean, this can be formulated in terms of signum (or heaviside function).

\[ \begin{align} h(x) &= \begin{cases} f(x) & x\geq a \\ g(x) & x\leq a \end{cases} \\ &=\begin{cases} f(x) & x> a \\ \frac{f(x)+g(x)}{2} & x=a \\ g(x) & x< a \end{cases} \\ &=\begin{cases} f(x) & x> a \\ \frac{f(x)}{2} & x=a \\ 0 & x< a \end{cases} + \begin{cases} 0 & x> a \\ \frac{g(x)}{2} & x=a \\ g(x) & x< a \end{cases} \\ &=f(x)\begin{cases} 1 & x> a \\ \frac{1}{2} & x=a \\ 0 & x< a \end{cases} + g(x)\begin{cases} 0 & x> a \\ \frac{1}{2} & x=a \\ 1 & x< a \end{cases} \\ &=f(x)\left(\frac{1}{2}+\frac{1}{2}\operatorname{sgn}(x-a)\right) + g(x)\left(\frac{1}{2}-\frac{1}{2}\operatorname{sgn}(x-a)\right) \\ \end{align}\]

Goal: Eliminate signum?

I have this little fun brilliant fact for 'simple' signum. That is: Given odd \(f:A\to\mathbb{R}, A=-A, f(x)\operatorname{sgn}(x)=f\left(|x|\right)\).

This is a quick little way to eliminate signum. So my first thought was - perhaps - if both \(f\) and \(g\) were odd functions, then we could simplify it. But we could only apply this when \(a=0\). And that does work, you end up with a nice \(\frac{1}{2}\left(f(x)+f(|x|)+g(x)-g(|x|)\right)\).

On our line of thinking with \(a=0\), my next thought was decomposing the functions - because as it turns out, we can simplify a piecewise equation with \(x^2\) and \(x\). That is, let \(f=f_{o}\cdot f_{a}\) and \(g=g_{o}\cdot g_{a}\), where \(f_{o}\) and \(g_{o}\) are odd function components of \(f\) and \(g\) respectively (not necessarily 0 for even functions), and \(f_{a}\) and \(g_{a}\) are the respective 'remaining' functions. From here we can actually simplify:

\[ h(x) = \frac{1}{2}f_{a}(x)\left(f_{o}(x)+f_{o}\left(|x|\right)\right)+\frac{1}{2}g_{a}(x)\left(g_{o}(x)-g_{o}\left(|x|\right)\right)\]

Interesting thing to note here is that for all odd functions:

\[ \frac{1}{2}\left(f_{o}(x)+f_{o}(|x|)\right) = \begin{cases} f_{o}(x) & x\geq 0 \\ 0 & x\leq 0 \end{cases} \]

\[ \frac{1}{2}\left(f_{o}(x)-f_{o}(|x|)\right) = \begin{cases} 0 & x\geq 0 \\ f_{o}(x) & x\leq 0 \end{cases} \]

With that being said, it's not clean at all. Hm. Bad line of thinking? Still doesn't answer why it's just so hard to find a nice form. We can't account for every function with this. Weird.

Signum identities

Also my next thought was - how could we potentially rewrite \(f(x)\operatorname{sgn}(x-a)\)?

Update: Well, \(f(x)\operatorname{sgn}(x-a) = f\left(|x-a|+a\operatorname{sgn}(x-a)\right)\), given odd \(f\). Consequently, \(f(x-a)\operatorname{sgn}(x-a) = f(|x-a|)\). Also interestingly, \(f(x-a)\operatorname{sgn}(x)=f\left(|x|-a\operatorname{sgn}(x)\right)\).

The latter two facts are consequences of the former fact, which is proven as follows. Suppose \(f\) is an odd function. Then:

\[ \begin{align} f(x)\operatorname{sgn}(x-a) &= \begin{cases} f(x) & x>a \\ 0 & x=a \\ -f(x) & x< a \end{cases} \\ &= f\left(\begin{cases} x & x>a \\ 0 & x=a \\ -x & x < a \end{cases}\right) \\ &= f\left(\begin{cases} x-a & x>a \\ 0 & x=a \\ -x+a & x < a \end{cases}+\begin{cases} a & x>a \\ 0 & x=a \\ -a & x < a \end{cases}\right) \\ &= f\left(|x-a|+a\operatorname{sgn}(x-a)\right) \\ \end{align}\]

New approach

So originally I did a subtitution with \(u=x-a\) on the original \(h(x)\) definition to help me focus - led me to this hypothesis: \(h(x)=f(\max(x,a))+g(\min(x,a))-f(a)\).

Turns out, this is indeed the case.

\[ \begin{align} h(x)&=\begin{cases} f(x) & x\geq a \\ g(x) & x\leq a \end{cases}\\ &=\begin{cases} f(x) & x\geq a \\ 0 & x< a \end{cases}+\begin{cases} 0 & x\geq a \\ g(x) & x< a \end{cases}\\ &=\begin{cases} f(x) & x\geq a \\ f(a) & x< a \end{cases}-\begin{cases} 0 & x\geq a \\ f(a) & x< a \end{cases}+\begin{cases} g(a) & x\geq a \\ g(x) & x< a \end{cases}-\begin{cases} g(a) & x\geq a \\ 0 & x< a \end{cases}\\ &=f\left(\begin{cases} x & x\geq a \\ a & x< a \end{cases}\right)+g\left(\begin{cases} a & x\geq a \\ x & x< a \end{cases}\right)-\begin{cases} g(a) & x\geq a \\ f(a) & x< a \end{cases}\\ &=f\left(\max(x,a)\right)+g\left(\min(x,a)\right)-\begin{cases} f(a) & x\geq a \\ f(a) & x< a \end{cases}\\ &=f\left(\max(x,a)\right)+g\left(\min(x,a)\right)-f(a)\\ &=f\left(\max(x,a)\right)+g\left(\min(x,a)\right)-g(a) \end{align}\]

Answer

It's not "difficult". It's just unintuitive; that is, the manipulation isn't particular obvious at first. With that being said, though, it should be a standard trick working with functions; these sorts of identities are important when building up and stripping down piecewise equations.