In class today we went over the Identity Criterion in Fitzpatrick's Advanced Calculus book. It says the following:
Let $I$ be an open interval and let the functions $f,g\colon I\to
\mathbb{R}$ be differentiable. Then these functions differ by a
constant iff $g'(x)=f'(x) \space \forall x \in I $.
We were told in class that its crucial that $I$ be an open interval. Otherwise the proposition does not hold. This is the example we were given to consider:
\begin{align}
f(x) = \left\{ \begin{array}{cc}
1 & \hspace{5mm} x<1 \\
2 & \hspace{5mm} x>2 \\
\end{array} \right.
\end{align}
\begin{align}
g(x) = \left\{ \begin{array}{cc}
3 & \hspace{5mm} x<1 \\
5 & \hspace{5mm} x>2 \\
\end{array} \right.
\end{align}
However, having had time to give it some thought, I still think that example satisfies the Identity Criterion. It doesn't differ by the same constant but it does differ by a constant. Am I thinking about this wrong?
Best Answer
By definition, a "constant" function is a function that takes the same value on all points of its domain. In your example, the function $h(x)=g(x)-f(x)$ is not constant, because (for instance) $h(0)=2$ but $h(3)=3$. A function like this $h$ which takes the same value near any particular point but may not take the same value on all points is instead called "locally constant".