Comparing the Sup and Inf of Two Functions

Question

If $f$ is a function $f$ : $D$ $\rightarrow$ $R$ one says that $f$ is bounded above (resp. bounded below, bounded) if the image of $D$ under $f$ i.e. $f(D)$ = {$f(x) : x\in D$} is bounded above (resp. bounded below, bounded). If $f$ is bounded above (resp. bounded below), then one denotes by sup $f$ the supremum of $f(D)$ (resp. by inf $f$ the infimum of $f(D)$).

Assume that two functions $f$:$D$$\rightarrow$ $R$ and $g:D\rightarrow $$R$ are bounded above.

(a) Prove that $f(x) \le g(x)$ for all $x \in D$ implies sup $f$ $\le$ sup $g$.

(b) Show that the converse it not true by providing a concrete counterexample.

(c) Prove that $f(x) \le g(y)$ for all $x, y \in D$ if and only if sup $f \le$ inf $g$.

Ok so this is the question, where there's three parts to it, and I feel as if that it's quite obvious, but I would like to see a nice proof for this question if possible. Really all I need to see is the proof for a), but if you want to do them all then go right ahead. I do understand why each of these are true, I just want to see what a good proof looks like. Thanks in advance.

Answer 1

(a) Let $y_0\in f(D)$. By definition of $f(D)$ we find $x_0$ such that $y_0 = f(x_0)$. Using the inequality we find $$y_1 = g(x_0) \ge f(x_0) = y_0$$ Thus for each element in $f(D)$ we have at least one larger element in $g(D)$. This means any sequence in $f(D)$, denoted by $(y^0_n)_n$ has a majorant sequence in $g(D)$, denoted by $(y^1_n)_n$. Now if we chose a sequence in $y^0$ to be a sequence convergent to $\sup f(D)$ (granted by the definition of supremum for sets), we can construct a sequence $y^1$ in $g(D)$ such that. $$y^1_n \ge y^0_n \quad\forall\ n$$ Using the property of the reals that every sequence has a convergent subsequence we can reindex $$\sup f(D) \leftarrow y^0_{n_l} \le y^1_{n_l} \to L$$ where we know that $L \ge \sup f(D)$. Finally we can conclude $\sup g(D) \ge L$ because it's the supremum and we have a convergent sequence in $g(D)$ wich converges to $L$, we find the desired result: $$\sup f = \sup f(D) \le L \le \sup g(D) = \sup g$$

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Properties used:

• There exists a sequence $x_n \to \sup X$, such that $(x_n) \subset X$
• Any sequence of real numbers has a convergent subsequence
• If a sequence $(x_n)\subset X$ is convergent, $\lim\limits_{n\to\infty} x_n \le \sup X$

Notice that most of the effort went into showing that $$\forall x\in X \exists y\in Y: x\le y \Rightarrow \sup X \le \sup Y$$ Wich is a property of the supremum for sets. If you have this property, all you need is the first step wich does not involve anything fancy. It's just the definition of a function and its image that guarantees us the property holds for $X=f(D), Y=g(D)$.

(b) Use $f(x) = 1$ and $g(x) = |x|$ on $D=(-1,1)$. $\sup f = \sup g$ but certainly $f(x) \not\le g(x)$ for any $x$ (this is so-to-speak a perfect example since we have strict inequality in the reverse order).

(c) @mfls' answer deals neatly with it so there's no need to lengthen this post.

Answer 2

Assume $f(x)\le g(y),\forall x,y\in D.$ We will argue by contradiction. So we suppose $\inf g<\sup f.$

Now, by definition, for any $\epsilon>0$ there exists $x,y\in D$ such that $f(x)+\epsilon\ge \sup f$ and $g(y)-\epsilon\le \inf g.$ Thus,

$$g(y)-\epsilon\le inf g<\sup f\le f(x)+\epsilon.$$

Now, consider $\epsilon=\frac{\sup f-\inf g}{2}.$ Thus,

$$g(y)-\frac{\sup f}{2}+\frac{\inf g}{2}\le \inf g<\sup f\le f(x)+\frac{\sup f}{2}-\frac{\inf g}{2},$$ from where,

$$0\le g(y)-f(x)\le \sup f-\inf g<0,$$ which is a contradiction.