Suppose that the temperature on the equator is a differentiable function $T(x)$ such that $T'(x)=0$ has only finitely many solutions. Prove that there are an equal number of local maxima and local minima.
$T(x)$ is $2\pi$-periodic as it is a temperature function on the equator. Hence we only need to consider the interval $[0,2\pi]$. Since $T(x)$ is differentiable, $T(x)$ is continuous. Hence, by the Extreme Value Theorem, $T(x)$ attains its maximum and minimum values on $[0,2\pi]$. Since $T(x)$ is differentiable on $[0,2\pi],$ by Fermat's Theorem, its maximum and minimum values either occur at endpoints or points where $T'(x)=0$. We now show that between any two consecutive local maxima, there is a local minima.
Suppose $x_1$ and $x_2$ are such that $T(x_1)$ and $T(x_2)$ are consecutive local maxima. Then $\exists \delta >0$ such that $x_1 < x < x_1 + \delta \Rightarrow T'(x_1) < 0$ and $\exists \delta_2 >0$ such that $x_2 -\delta < x < x_2\Rightarrow T'(x_2) >0.$ Pick $c\in (x_1, x_1 + \delta)$ and $c_2 \in (x_2-\delta, x_2)$ such that $c_2 > c$. Consider the interval $[c,c_2]$. Since $T(x)$ is continuous on this interval, by the Extreme Value Theorem, it must attain its maximum and minimum values on that interval. We show that the minimum value is not attained at the endpoints. Since $T'(c) < 0,$ we have that $\lim\limits_{h\to 0^+}\dfrac{T(c+h)-T(c)}{h} < 0\Rightarrow T(c+h) < T(c)$, and so $T(c)$ is not a minimum. Similarly, $T'(c_2) >0$, so $\lim\limits_{h\to 0^-}\dfrac{T(c_2+h)-T(c)}{h} >0\Rightarrow T(c_2+h)-T(c) < 0$, so $T(c_2)$ is not a minimum either. Therefore, the minimum is attained in $(c,c_2)$. By similar logic, we can show that there is a maximum between two consecutive local minima. Hence there must only be one local minimum between two consecutive local maxima (otherwise those maxima would not be consecutive). Hence if $T(x)$ only attains one local maximum in $[0,2\pi]$, it only attains one local minimum. As well, if we let $n$ be the number of local maxima in $[0,2\pi]$ and $m$ be the number of local minima, then $|n-m| \leq 1$ as otherwise there will be two consecutive local maxima without a local minima in between or vice versa.
Is there a shorter proof of this?
Isn't there a shorter proof of this?