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Let A be the set A= { $x^{2}+x\cdot y^{2}$ : $-2 \leq x \leq 1, -1\leq y \leq 1$ }

Determine wether the set A, subset of the reals is bounded above, below both or neither. Rigorously justify your answer.

Notes : I'm not allowed to use limits or calculus . So far I've found A is bounded above and below simply by minimising and maximising the values of x and y, however I don't think this is rigorous enough. Is there any more formal proof showing that the set is bounded above and below ? Any help is much appreciated.

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    $\begingroup$ Finding the extremes of the function $f(x,y)=x^2+xy^2$ on the compact domain $[-2,1]\times [-1,1]$ is perfectly rigorous. But the answer by bat_of_doom is the easy way. $\endgroup$ – Mathematician 42 Feb 5 '17 at 9:30
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    $\begingroup$ The set $A$ is the image of a continuous function (i.e., $(x,y)\mapsto x^2+xy^2$) on a connected compact set (i.e., $[-2,1]\times[-1,1]$). Therefore, $A$ is also a connected compact set. Being a subset of $\mathbb{R}$, $A$ is a closed finite interval. In fact, it is easy to see that $A=\left[-\frac14,4\right]$. $\endgroup$ – Batominovski Feb 5 '17 at 9:36
  • $\begingroup$ @Batominovski Thank you :) $\endgroup$ – Anonymous Feb 5 '17 at 10:53
  • $\begingroup$ @Mathematician42 Thank you ! :) $\endgroup$ – Anonymous Feb 5 '17 at 10:53
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You only need a lower and upper bound, not the greatest and least respectively.For that, $$|x^2+xy^2|\leq |x|^2+|x||y|^2\leq4+2=6$$, and you are done.

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    $\begingroup$ To the OP: Note that you do not need an accurate estimate just to show boundedness. When trying to stay below $\infty$ and above $-\infty,$ any bounds will do as long as they're finite. $\endgroup$ – DanielWainfleet Feb 5 '17 at 9:56
  • $\begingroup$ Thank you, and by that you can show that the set A has a range less than or equal to 6 right ? $\endgroup$ – Anonymous Feb 5 '17 at 10:51
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    $\begingroup$ @Sonia Also, greater than equal to -6. (Mind the modulus.) So, -6 serves as a lower bound and +6 as upper. $\endgroup$ – bat_of_doom Feb 5 '17 at 11:39
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A pedestrian approach:

1) Lower bound of $A = x^2 + xy^2$ , where $-2 \leq x \leq 1$ and $-1\leq y \leq 1$.

Let $ \alpha: = y^2$, then $0\leq \alpha \leq 1$.

Consider:

A($\alpha$ )= $x^2$ + $\alpha$ x =

= $( x + \alpha /2)^2$ - $\alpha^2$ / 4.

This is a family of parabolas, $\alpha$ the parameter, with minimum = - $\alpha ^2$ / 4, $\\$ at ($-\alpha$ /2, -$\alpha ^2 $ / 4).

Choose $\alpha$ = 1 (y= +1 or y = -1) to find the curve with the smallest minimum of the family of $\alpha$ curves.

Finally:

$$A(\alpha) = (x+\alpha /2)^2 - \alpha ^2 / 4 \geq \\ -\alpha ^2 / 4 \geq - 1/4 .$$

2) Upper bound for A in the given x,y interval, a bit of patchwork.

Consider the intervals - 2 $\leq$ x $\leq$ 0, and -1 $\leq$ y $ \leq$ 1 , I.e. $ \alpha$ $\geq$ 0.

$A(\alpha) = x^2 + \alpha x \leq A(\alpha=0) = x^2$ , recall that $\alpha \geq 0$, so the left hand side is smaller than $ x^2$ for negative x.

A($\alpha$) $\leq x^2 $, for - 2 $\leq$ x $\leq 0 $,

Setting x = - 2 gives A($\alpha$) $\leq$ A($\alpha$ = 0) = 4 in the intervals considered.

Almost done.

Now consider the remaining interval 0 $\leq$ x $\leq$ 1, $\alpha$ as before.

$$A(\alpha) = x^2 + \alpha x \leq A(\alpha = 1) = \\ = x^2 + 1x ,$$ where all the x terms are positive (or zero) and $\alpha = 1$.

The maximum for A($\alpha=1$) = 2 at x=1.

For the complete interval -2 $\leq$ x $\leq$ 1, with $\alpha$ (y) in the given range:

Maximum = max( 2,4) = 4 .

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Your $x\in[-2,1]$ and $y\in[-1,1]$ is a compact set. $x^2+xy^2$ is a continuous function so that both minimizing and maximizing $x^2+xy^2$ over the set has a solution. Set lower and upper bound equal to $x^2+xy^2$ evaluated at the minimum and maximum respectively.

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