If $\lim\limits_{x\rightarrow{a}}f(x)=\lim\limits_{x\rightarrow{a}}\lfloor f(x)\rfloor,$ prove that...

Question

If $\lim\limits_{x\rightarrow{a}}f(x)=\lim\limits_{x\rightarrow{a}}\lfloor f(x)\rfloor,$ prove that $f(x)$ has a local minimum at $x=a$. It is given that $f(x)$ is a non constant, continuous function.

Here, $\lfloor\cdot\rfloor$ is the greatest integer function.

$$Let \space RHL=\lim\limits_{h\rightarrow{0}}f(a+h)=\lim\limits_{h\rightarrow{0}}\lfloor f(a+h)\rfloor=k, k \in I$$ $$LHL=\lim\limits_{h\rightarrow{0}}f(a-h)=\lim\limits_{h\rightarrow{0}}\lfloor f(a-h)\rfloor =k$$

How to proceed further?

Answer 1

As $f$ is supposed to be continuous and $\lfloor f(x)\rfloor$ takes integer values, $f(a)=\lim\limits_{x \to a} f(x)$ is an integer $k \in \mathbb Z$.

If $f$ doesn’t have a minimum at $a$, then it exists a sequence $\{a_n\}$ converging to $a$ with $f(a_n) \lt k$ for all $n \in \mathbb N$. But then $\lfloor f(a_n) \rfloor \le k-1$ for all $n \in \mathbb N$. Which implies the contradiction $$f(a)= \lim\limits_{n \to \infty} \lfloor f(a_n)\rfloor \le k-1 \lt k =f(a)$$

Answer 2

Let me try to clarify the missing point in the accepted answer

$$\lim_{x \to a} f(x)=\lim_{x \to a} \lfloor f(x) \rfloor = f(a) \in \mathbb{Z}$$

$$\lim_{x \to a} \lfloor f(x) \rfloor=\lim_{x \to a} f(x)-\{f(x)\} = \lim_{x \to a} f(x)-\lim_{x \to a} \{f(x)\}$$

providing that both limits exist.

We know only about $\lim\limits_{x \to a} f(x)$ so we have that $\lim\limits_{x \to a} \{f(x)\}$ must exist too since $\lim\limits_{x \to a} \lfloor f(x) \rfloor$ exists. So

$$\lim_{x \to a} \lfloor f(x) \rfloor=f(a)=\lim_{x \to a} f(x)-\{f(x)\} = f(a) -\lim_{x \to a} \{f(x)\}$$

or

$$\lim_{x \to a} \{f(x)\}=0$$

In general, $f(a) \in \mathbb{Z}$ implies $$\lim_{x \to a} \{f(x)\} \in \{0,1\}$$ providing this limit exists, and we know that it does.

This strictly defines how $\{f(x)\}$ may approach the limit around the same whole number. If it is from bellow, it is $\lim\limits_{x \to a} \{f(x)\} =1$, if it is from above, it is $\lim\limits_{x \to a} \{f(x)\} =0$.

It cannot approach both ways as that would render the limit undefined since we would jump from $0$ to $1$ while we approach $a$. (And this is actually the reason it has the minimum.)

Because of that (that $\lim\limits_{x \to a} \{f(x)\} =0$ and that it approaches $0$ from above) there is a vicinity of $a$, $\delta > 0$ for which:

$$ f(a+x) \geq f(a), |x|<\delta$$

This is making $f(a)$ a local minimum.

The key point is:

$\lim\limits_{x \to a} \{f(x)\}$ cannot approach $0$ both ways, from bellow and from above, as that would render the limit undefined.

In the same manner

$$\lim_{x \to a} f(x)=\lim_{x \to a} \lceil f(x) \rceil$$

makes $f(a)$ a local maximum.

In both cases, we are just talking only about this specific case that gives a min or a max respectively:

$$\lim_{x \to a} \{f(x)\} \in \{0,1\}$$