# Suppose a continuous function attains its minimum, prove that the function is not injective

Suppose a continuous function $$f: (0,2) \to \mathbb R$$ attains its minimum at $$x_0 \in (0,2)$$, prove that the function is not injective.

We need to show there are some $$a$$ and $$b$$ such that $$f(a)=f(b)$$. I think we can first notice that since $$f(x_0)$$ is minimal, for any $$x$$, we have: $$f(x_0) \leq f(x)$$

Now what the statement is saying is that there must also be some $$x_1$$ such that $$f(x_0) \geq f(x_1)$$. I cannot use the extreme value theorem because we do not have a closed interval here. How could I proceed?

• what are the other condition? is not the function continuous? – Bijayan Ray Jan 27 at 10:11

Since $$x_0$$ lies in the open interval $$(0,2)$$, there exist $$a,b$$ with $$0 < a < x_0 < b < 2$$.

If $$f(a)=f(b)$$ or $$f(a)=f(x_0)$$ or $$f(b)=f(x_0)$$, we are done.

If $$f(a) < f(b)$$, we have $$f(x_0) < f(a) < f(b)$$. By the intermediate value theorem for the continuous function $$f$$ and the interval $$[x_0,b]$$, there exists an $$x_1$$ with $$x_0 < x_1 < b$$ and $$f(x_1)=f(a)$$. By the given inequalities, we have $$a < x_1$$.

If $$f(a) > f(b)$$, we have $$f(a) > f(b) > f(x_0)$$. Again by the intermediate value theorem for the continuous function $$f$$ and the interval $$[a,x_0]$$, there exists an $$x_1$$ with $$a < x_1 < x_0$$ and $$f(x_1)=f(b)$$. By the given inequalities, we have $$b > x_1$$.

• The IVT works in mysterious ways. It took me a while to figure out how subtle this is. – Wesley Strik Jan 27 at 11:08

It seems you can make the interval closed by taking the interval between $$\frac {x_0}{2}$$ and $$\frac {x_0 + 2}{2}$$ and then apply the extreme value theorem if is continuous on [0,2].

Let the minimum be attained at $$x_0$$. Consider the maximum of $$f$$ on $$[\frac{x_0}{2}, x_0]$$, and consider the maximum of $$f$$ on $$[x_0, \frac{x_0+2}{2}]$$. What can you say about $$f$$'s values on these two intervals?