Proving something is a lower or upper bound.

Question

Say we have a set containing the elements:

{$2x^2 + 2/x^2$: $ 1/2 < x < 2$}.

I intuitively (and graphically) know that the supremum and the infimum are 8.5 (and that they're not maximum nor minimum), but how do I actually prove that they're the supremum and infimum. How do I go about showing they're larger than/smaller than all elements in the function respectively? Nothing too advanced please, still learning the basics of analysis. Thanks!

Answer 1

Here you are essentially asking for the range of the function $f:(\frac{1}{2},2) \to \mathbb{R}$ defined by $f(x)=2x^2+2x^{-2}$. We can compute the derivative: $f'(x)=4x-4x^{-3}$. We see that $f'(x)=0$ implies $x=1$. Since $f$ is continuous, we can compute the limit at the boundary by substitution:

\begin{equation*} \lim_{x \to \frac{1}{2}^+} f(x) = \frac{17}{2}; \, \lim_{x \to 2^-} f(x) = \frac{17}{2} \end{equation*}

Furthermore, we have $f(1)=4$. We conclude that the given set has infinum $4$, which it attains, and suprenum $\frac{17}{2}$ which it does not attain.

Answer 2

If you really want to do it using only definitions you could try this.

Let $S = \{2x^2+2x^{-2}\;|\;x \in (1/2,2)\}$, the infimum is the difficult part. First remark that $4$ is a L.B of $S$, because $4 \leq 2x^2+2x^{-2}$ for all $x \in (1/2,2)$. Then claim that $\inf(S) = 4$, so we need only show that any other lower bound of $S$ say $a$ has $a \leq 4$. So because $a$ is a L.B, $a \leq 2x^2+2x^{-2}$ for all $x \in (1/2,2)$. Take $x = 1$, then $a \leq 2(1)^2+2(1)^{-2}=4$, hence $\inf(S) =4$.

A similar argument would be used for the supremum.

You could also just say that because $4$ is the minimum of the set, it is automatically the infimum.

Answer 3

Hints.

For the minimum consider $2x^2+2/x^2=2(x-1/x)^2+4$.

For the maximum, suppose $a>b\ge 1.$ Then $(a+1/a)-(b+1/b)=$ $(a-b)-(a-b)/ab=$ $(a-b)(ab-1)/ab>0.$

And if $0<a'<b'\le 1,$ let $a=1/a'$ and $b=1/b'.$ Then $a>b\ge 1$ and $(a'+1/a')-(b'+1/b')= (a+1/a)-(b+1/b).$