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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?

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  • $\begingroup$ what are the other condition? is not the function continuous? $\endgroup$ – Bijayan Ray Jan 27 at 10:11
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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 $b > x_1$.

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

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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?

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