Can the supremum be equivalent to a function?

Question

Let $f : \Bbb{R} \to \Bbb{R}$ be a bounded function and $h(y) \equiv \sup_{x\ge y} f(x)$ for all $y \in \Bbb{R}$. Show that $h$ is decreasing.

In this question, how can $h(y)$ be equivalent to the supremum and at the same, be decreasing? Is $h(y)$ a function? Moreover, how can we look for the supremum of $f(x)$ when $x$ is bigger than $y$? In such a case, a supremum $h(y)$ would not exist.
Is the question saying that as $x$ increases, the supremum (as a point) will decrease? Does it mean that the function as a whole is strictly decreasing?

I would really appreciate any clarification on what the question is actually asking for. I do not have a good math background and it's my first time covering these topics.

Answer 1

Some examples may help illustrate how this is all possible.

Example 1: $f(x) = -\arctan(x)$.

This function is decreasing. So, given any $y$, and any $x \ge y$, we know that $f(x) \le f(y)$. Illustrated above is what happens when we take $y = 1$ (and so $f(y) = -\frac{\pi}{4}$). For $x \ge 1$, the function value only decreases below $-\frac{\pi}{4}$. The supremum of all these function values is going to be $-\frac{\pi}{4}$. That is, $h(1) = -\frac{\pi}{4}$.

This works for any $y$. In this case, $h(y) = f(y)$ for all $y$, since $f$ is decreasing. What about when $f$ is not decreasing?

Example 2: $f(x) = \frac{1 - x^2}{1 + x^2}$.

The first picture illustrates $h(-3) = 1$, and the second illustrates $h(2) = -\frac{3}{5}$.

In the first picture, $f(x)$ achieves a maximum of $f(x) = 1$, over $x \ge -3$, at $x = 0$. The supremum of all the function values $f(x)$, for $x \ge -3$, will therefore be $1$.

In the second picture, $f(x)$ is decreasing once again over $x \ge 2$. So, the supremum is achieved at $x = 2$, so $h(2) = f(2) = -\frac{3}{5}$.

Note that, for $y \le 0$, the function $f(x)$ will achieve its maximum of $1$, over $x \ge y$. So, for $y \le 0$, we will have $h(y) = 1$ constantly. But, as soon as $y > 0$, and the function begins decreasing, we will have $h(y) = f(y)$ as we did in the first example. So, in this case: $$h(y) = \begin{cases} \frac{1 - y^2}{1 + y^2} & \text{if } y > 0 \\ 1 & \text{if } y \le 0.\end{cases}$$

Example 3: $f(x) = \sin(x)$

The above illustrates that $h(-\pi) = 1$. In fact, $h(y) = 1$ for all $y$, since the $\sin$ function achieves a maximum of $1$ past any $y$ you choose. No matter how much of the $x$ axis you cut off, there will be a point to the right where $\sin(x) = 1$. Thus, $h(y) = 1$ for all $y$. (Note: like the last example, this is non-strictly decreasing.)

Anyway, that should hopefully illustrate how this function $h$ works. I'll leave the rest to you.

Answer 2

For each real number $y$ there is a half-open interval $I_{y} := [y, \infty) = \{x : y \leq x\}$. These intervals are nested inward, in the sense that $I_{y'} \subseteq I_{y}$ if and only if $y \leq y'$.

Let $J_{y} = f(I_{y})$ denote the set of values $f$ achieves on $I_{y}$. These sets are also nested inward; why?

Finally, we've defined $h(y) = \sup J_{y}$. Can you take it from here?