Exercise 1.6.4 (Uniqueness of antiderivatives up to constants)

Proof. The if part follows directly from the definition. Now suppose that $F^{\prime }=0$, and suppose for the sake of contradiction that $F$ is not constant. Then we can find two distinct points $x_1,x_2\in [a,b]$ such that $F(x_1)\ne F(x_2)$. By mean value theorem, there exists a $x\in (x_1,x_2)$ such that $F^{\prime }(x)=\frac{F(x_2)-F(x_1)}{x_2-x_1}\ne 0$ - a contradiction.

Now we return to the exercise.

• Suppose that $F(x)=G(x)+C$. Then, the assertion follows directly from the fact that the derivative is additive (cf. Analysis I, Theorem 10.1.13), and the derivative of a constant is zero, i.e.,

  $$F(x)=G(x)+C⟹\frac{dF}{dx}=\frac{d(G+C)}{dx}=\frac{dG}{dx}+\frac{dC}{dx}=\frac{dG}{dx}$$

• Now suppose that $F^{\prime }(x)=G^{\prime }(x)$ and pick an arbitrary $x\in [a,b]$. Define $H:=F-G$, we then have by theorem assertion $H^{\prime }=F^{\prime }-G^{\prime }=0$. By the corollary to the Mean Value Theorem, $H$ must be a constant function, i.e., $\forall <!--FUNCTION\; APPLICATION-->x\in [a,b]:H(x)=F(x)-G(x)=C$ for some $C\in ℝ$, as desired.