The identity criterion

Question

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?

Answer 1

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".

Answer 2

The constant should be the same constant.

If f'(x)=g'(x) on some interval then we have $$f(x)= f(x_0)+\int _{x_0}^x f'(t)dt=$$

$$ f(x_0)+\int _{x_0}^x g'(t)dt=$$

$$ f(x_0)+ g(x)-g(x_0) $$

$$\implies f(x)-g(x) = f(x_0)-g(x_0)$$

That is $f(x)-g(x)$ is a constant and the constant is the same for all x in your interval.