Why does $\displaystyle f'(x)=0 \; \forall \; x\in\mathbb{R}\implies f(x)=c $?

Printable View

- Jun 22nd 2010, 09:59 AMmathman88Simple question
Why does $\displaystyle f'(x)=0 \; \forall \; x\in\mathbb{R}\implies f(x)=c $?

- Jun 22nd 2010, 10:16 AMAckbeet
Integrate both sides. What do you get?

- Jun 22nd 2010, 11:59 AMmathman88
I understand that, but does that imply that's the unique solution?

- Jun 22nd 2010, 12:01 PMAckbeet
Well, you have an infinite family of solutions depending on your initial condition. However, the fact that the RHS of your DE, namely 0, is Lipschitz and continuous, is enough to guarantee the uniqueness of the solution to an initial value problem (an initial value problem is your original DE plus an initial condition to determine the constant of integration). Does that make sense?

- Jun 22nd 2010, 12:23 PMmathman88
Why does Lipschitz continuity guarantee uniqueness of the solution (up to a constant)?

- Jun 22nd 2010, 12:31 PMAckbeet
Because of the Picard-Lindelof theorem. If you're curious, by all means investigate the proof of that theorem.

- Jun 26th 2010, 02:59 PMHallsofIvy
For this specific question, you don't need the full "Picard-Lindelof", just the mean value theorem.

Let $\displaystyle x_0$ and $\displaystyle x_1$ be two distinct values of x: $\displaystyle x_0\ne x_1$.

By the mean value theorem, $\displaystyle \frac{f(x_1)- f(x_0)}{x_1- x_0}= f'(c)$ where c is some number between $\displaystyle x_0$ and $\displaystyle x_1$. If f'(x)= 0 for all x, then f'(x)= 0 so $\displaystyle f(x_1)- f(x_0)= 0$ and $\displaystyle f(x_1)= f(x_0)$.