Shorter way to prove $|\underset{x \in A}{\text{inf}}f(x)-\underset{x \in A}{\text{inf}}g(x)|\leq \underset{x \in A}{\text{sup}} |f(x)-g(x)|$

Question

Here is a result I found in a step of a proof (not even a lemma, and I can't even find its justification):

$$|\underset{x \in A}{\text{inf}}f(x)-\underset{x \in A}{\text{inf}}g(x)|\leq \underset{x \in A}{\text{sup}} |f(x)-g(x)|$$

Here're my questions:

1. Is my proof correct? (I've attached it below)
2. Is there a more intuitive way to do it? What's the intuition behind? Can I understand the result without the concept of taking limit?

My attempt to prove it:

Denote $sup= \underset{x \in A}{\text{sup}} |f(x)-g(x)|$ (assume it's finite), and assume both $\text{inf} f(A)$ and $\text{inf} g(A)$ are real.

For all $x \in A$, $|f(x)-g(x)| \leq sup$. By the definition of infimum, we can find a sequence $(f(x_n))$, where $x_n \in A$, such that

$$\lim_{n \to \infty }f(x_n)=\underset{x \in A}{\text{inf}}f(x):= inf_f$$

So for all $x \in A$ we have:

$$|inf_f-g(x)|\leq sup$$

Similarly for $g$ we get $|inf_f-inf_g|\leq sup$.

Am I correct?

Answer 1

As I pointed in the comments, your proof is not correct.

Assume we don't have $\inf f(A)=\inf g(A)=-\infty$ (otherwise the sentence makes no sense) and we have $sup:=\underset{x \in A}{\sup} |f(x)-g(x)|<+\infty$ (otherwise the sentence is obvious).

The intuition is to think to the graph of $g$ as being inside a tube of radius $sup$ around the graph of $f$, i.e. $$\forall x\in A\quad f(x)-sup\le g(x)\le f(x)+sup.$$ Take the infimum (over $x\in A$) of the three terms and you are done.

Answer 2

I do have a more properties-based way to answer it.

Now to show that $|B| \leq C$, we can show that $B\leq C$ and $-B \leq C.$

So we first show that $\inf f(x) - \inf g(x) \leq \sup |f(x) - g(x)|.$

It follows by the following: $$\begin{align}\inf f(x) - \inf g(x)&\leq \inf f(x) + \sup -g(x)\\&= \inf f(x) + \sup (f(x) - g(x) - f(x))\\&\leq \inf f(x) + \sup (f(x) - g(x)) + \sup -f(x)\\&= \inf f(x) + \sup (f(x) - g(x)) - \inf f(x)\\&= \sup (f(x) - g(x))\\&\leq \sup |f(x) - g(x)|\end{align}$$

The $-B \leq C$ case is proved if you switch $f$ and $g.$