I'm self learning real analysis from a book and in one of the exercise I struggle to understand the correction of the second point. Here is the instruction of the exercise :

$f$ is an injective function form $I$ to $\mathbb R$, continue on $I$:

  1. Assume that f is not strictly monotonic on I. Show that there exist 4 real number a,b,x,y in I such $x <y$ and $f(x)>f(y)$, $a<b$ and $f(a)<f(b)$.

This was straight forward, I write what it mean in term of quantificator.

$$ \exists (x,y)\in I^2 (x \neq y\Rightarrow f(x)\geq f(y)) \\\text{and}\\ \exists (a,b)\in I^2 (a \neq b\Rightarrow f(a)\leq f(b))$$ and because f is an injectiv function we can replace the inequality signs by strict one.

  1. Consider the map: $$\phi(x): \lambda \in [0,1] \mapsto f((1-\lambda)b+\lambda y)-f((1-\lambda)a+\lambda x)$$ The goal is to show that there exists $\lambda_0$ such that $\phi(\lambda_0)=0$.

Here is where I struggle. I understand that $(1-\lambda)b+\lambda y$ lie in between $b$ and $y$ similarly that $(1-\lambda)a+\lambda x)$ lie in between $a$ and $x$ and therefore those two number belong to $I$. However on the solution of the exercise they say that $\phi(a)=\phi(b)=0$ and this is the part that I don't understand. Then,they used the mean value theorem to conclude that $\lambda_0$ exist.

  1. This was straight forward but in case someone is interested in the exercise. Show that $\phi(\lambda_0)=0$ is a contradiction with f being injective. Then conclude that an injective function for $I$ to $\Bbb R$ is strictly monotonic.

1 Answer 1


Note that$$\phi(0)=f(b)-f(a)>0\quad\text{and}\quad\phi(1)=f(y)-f(x)<0.$$So, since $\phi$ is continuous, it must be equal to $0$ somewhere between $0$ and $1$.

  • $\begingroup$ It seems so obvious now. Thanks a lot José $\endgroup$ Commented Jan 6, 2022 at 12:11

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