# Let $A$ be a subset of $mathbb{R}$ such that $A$ is bounded below with inf $A = L > 0$.

Mathematics Asked by Outlier on July 26, 2020

Let $$A$$ be a subset of $$mathbb{R}$$ such that $$A$$ is bounded below with inf $$A = L > 0$$. Prove that if $$a,b < 0$$, then the set $$B = {ax^2 + bx} : x in A$$ is bounded above and sup $$B = aL^2 + bL$$.

Suppose $$k in B$$. Then $$k = ax^2 + bx$$ for some $$x in A$$.

inf $$A = L > 0$$, $$x geq L$$ for all $$x in A$$. Hence $$k leq aL^2 + bL$$ since $$a,b < 0$$ for all $$k in B$$. Therefore $$B$$ is bounded above.

Next we show that sup $$B = aL^2 + bL$$.

Since sup $$B$$ is the least upper bound for $$B$$, sup $$B leq aL^2 + bL$$.

Let $$epsilon > 0$$. By the characterization of supremum, $$exists$$ $$k = ax^2 + bx in B, x in A, a,b < 0$$ such that $$k > aL^2 + bL – epsilon$$.

This implies that $$aL^2 + bL – epsilon < aL^2 + bL leq$$ sup $$B$$.

Hence $$aL^2 + bL <$$ sup $$B + epsilon$$. Since this holds for every $$epsilon > 0$$, we have sup $$B = aL^2 + bL$$.

Is the proof above correct?

Since, $$inf A = L > 0$$ , $$A$$ is the set of positive reals, $$(-a)x^2 + (-b)x > (-a)L^2 + (-b)L > 0$$
implies $$ax^2 + bx < aL^2 + bL < 0$$
So, $$aL^2 + bL$$ is a upper bound for $$B$$. ( $$B$$ is the set of negative reals)
Now, we have to show that $$aL^2 + bL$$ is a least upper bound.
If possible let there is some upper bound $$aM^2 + bM$$ of $$B$$ such that $$aM^2 + bM < aL^2 + bL < 0$$
this implies $$(-a)M^2 + (-b)M > (-a)L^2 + (-b)L > 0$$
from these we can deduce that $$M > L$$ and $$M$$ is a lower bound of $$A$$.(because if $$M$$ is not a lower bound of $$A$$, then $$aM^2 + bM$$ can't be a upper bound of $$B$$).
So, clearly from $$M > L$$ , we get a contradiction to the fact that $$L$$ is greatest lower bound of $$A$$.
So, our assumption is wrong.
Hence, $$aL^2 + bL = sup B$$

Answered by Subhajit on July 26, 2020

No it's not correct. This inequality, $$k > aL^2 + bL - epsilon$$, cannot be derived for all $$epsilon$$. The respective inequality can be only derived for $$sup B$$, since that is the supreme.

You can assume thou that if $$aL^2 + bL > sup B$$ then by setting $$epsilon_0= aL^2+bL-sup B>0$$, you have that, there exist $$x_0$$ such that, $$0>aL^2+bL geq ax_0^2+bx_0geq sup B-epsilon_0=-(aL^2+bL)> 0$$, which is absurd

Answered by random certainty on July 26, 2020