“Analogue” of the Extreme Value Theorem to prove non-existence of max. and min. points

In general, suppose that we are interested in the maximum and minimum properties of the derivative $f'$ of a particular function $f$, and in the first part, we further assume continuity of $f'$ on an interval $I$. Then by a simple application of the Extreme Value Theorem (Baby Rudin 4.16):

Suppose $f$ is a continuous real function on a compact metric space $X$, and

$$M = \sup_{p \ \in \ X} f(p), \quad m = \inf_{p \ \in \ X} f(p)$$

Then $\exists \ p, q \in X$ s.t. $f(p) = M$ and $f(q) = m$.

we can ascertain the existence of a local maximum and minimum. However, now suppose that at first the first derivative is discontinuous somewhere on the interval $I$, and the problem now is a matter of trying to prove or disprove existence of such extreme values. Does there exist any analogue to the EVT, but for discontinuous functions?

I understand that one might recommend to take the second-order derivative of $f$ and attempt to see if $\exists \ x \in I$ s.t. $f''(x) = 0$, but for the purposes of this question, let us assume that $f$ is not twice differentiable, thus forcing us to merely work with the continuity (or discontinuity) property of $f'$.

• No, as far as I know not. Even if you know that $f'$ is bounded you may or may not attend the extreme values. – crankk Jul 6 '17 at 6:53

However, now suppose that at first the first derivative is discontinuous somewhere on the interval I

The assumption is that $f$ is continuous. Therefore $f$ can be nowhere differentiable and the theorem still holds. The proof also just use the continuity of $f$ and the compactness on the metric space $X$.

By definition of $M$, there exists a sequence $(x_n)_n\subset X$ such that $f(x_n)\to M$. Since $X$ is compact, we get a convergent subsequence $(x_{n_k})_k$ with $x_{n_k}\to x\in X$ and still $f(x_{n_k})\to M$ for $k\to\infty$. Since $f$ is continuous we get $f(x_{n_k})\to f(x)=M$. Especially $M<\infty$. Same goes for $m$.

You might consider $X=[0,1]$ and $f:X\to \mathbb{R}$ defined by $f(x)=|x|$. Then you have $m=0$ and $M=1$ and $f(0)=0$ although "$f'$ is discontinuous at $0$" and $f(\pm1)=1$ altough $f'(\pm1)=\pm1\neq 0$.

Or a little bit more pathological: $g:X\to \mathbb{R}$ with $$g(x)=\begin{cases}1 & x\in\mathbb{Q}\\ 0 & x\in\mathbb{R}\setminus\mathbb{Q}\end{cases}$$ There is still $m=0$ and $M=1$ and $g$ is nowhere differentiable but $g\left(\frac{\pi}4\right)=0$ and $g(0)=1$.

Well, at first, any kind of differentiability implies continuity, so, there is no doubt that extreme values do exist and the theorem "works".

Let's look at discontinuous functions, now. For instance, if $X=[-1,1]$ with it regular metric and let $f$ be the following, discontinuous, function: $$f(x)=\left\{\begin{array}{ll}\frac{1}{x} & \mbox{if x\neq0}\\0 & \mbox{else}\end{array}\right.$$ It is obvious that $f$ has no finite extreme values, so the theorem does not stand.

A more "soft" case would be that of (now $X=[0,1]$): $$f(x)=\left\{\begin{array}{ll}x & \mbox{if x\neq1}\\0 & \mbox{if x=1}\end{array}\right.$$

With the notation given in the question, $M=1$ but there's no such $p\in X$ such that $f(p)=1=M$, so, again the theorem does not stand without continuity.

However, it could be possible if you "enhance" your space $X$. For instance, if $X$ is ordered - e.g. $(\mathbb{R},<)$, then, every real monotonous function $f$ on $X$ satisfies the result of the theorem. Let's prove it:

Let $f:X\to\mathbb{R}$ be a monotonous function, let's suppose it is increasing, and let $$M=\sup_{x\in X}f(x)\mbox{ and }m=\inf_{x\in X}f(x)$$ Then, since $X$ is compact, it is evident it is bounded and, hence, there exists $p=\sup X$ and it is finite. Moreover, by the characterization of $\sup X$, we can construct a sequence $p_n$ which converges to $p$. Again due to compactness, we have that $p\in X$, and, due to monotonicity, $f(x)\leq f(p)$ for every $x\in X$.

In the same way we can show than $f(q)\leq f(x)$ for every $x\in X$, where $q=\inf X$.

So, to sum up, every hypothesis about differentiability, regardless the continuity of $f'$, takes us back to the case where $f$ is continuousa and any declination of continuity can ruin - as shown - the result of the theorem vastly. However, one can have the same result for discontinuous funtions, if the properties of $X$ are altered or "enhanced" a "little bit" (with whatever may that "little bit" mean).

In general if you want existence of the minimum the lowest regularity assumption i guess is lower semicontinuity of the function (see the direct method of calculus of variation). That is \begin{equation*} \liminf_{x\rightarrow x_0}f(x)\geq f(x_0)\quad \mbox{ at any } x_0 \end{equation*} And if you want existence of maxima the lowest regularity assumption i guess is to ask for upper semicontinuity that is \begin{equation*} \limsup_{x\rightarrow x_0}f(x)\leq f(x_0)\quad \mbox{ at any } x_0 \end{equation*} But is easy to show that a function that is both upper and lower semicontinuous at any point it is continuous as \begin{equation*} f(x_0)\leq\liminf_{x\rightarrow x_0}f(x)\leq\limsup_{x\rightarrow x_0} f(x)\leq f(x_0). \end{equation*}