# max/min and compactness

if $$f(x)=\frac{1}{1+x^2}$$ and $$-\infty, show that $$f$$ attains a $$max$$ but does not attain a $$min$$.
what I ve learnt so far is that :
if $$f$$ is a real valued function from a compact metric space, then $$f$$ attains a max and min at some point of $$M$$.
if $$f$$ is a real valued function from a closed bounded set, then $$f$$ attains a max and min at some point of that set.

I believe I must be able to apply any of these two statements to analyse such questions. right?
Now since $$-\infty and not closed in $$R$$, so I ignore the second statement.
but for the first statement since $$-\infty represents real numbers $$R$$, and we know $$R^n$$ is not compact, so the first statement does not apply!
Well this is an excercise after those two statements. how are they going to help then?

• it is the exact text of the exercise in the book – BesMath Mar 27 '20 at 21:19
• You are correct. The statements in no way apply. But that's fine. You solve it by other means. This isn't a question about compactness. It's about bounded sets and sups and infs. Now $x^2 \ge 0$ so $1+x^2 \ge 1$ so $f(x) \le 1$. So $1$ is an upper bound. And note $f(x) > 0$ so $0$ is a lower bound. Can you go on. – fleablood Mar 28 '20 at 0:08
• "Well this is an excercise after those two statements. " This could be a set up for the next excercise.... – fleablood Mar 28 '20 at 0:09
• I see thank you. – BesMath Mar 28 '20 at 0:11

To maximize $$f(x)$$, you need to minimize the denominator. The minimum value of the denominator is 1 because $$x^2 \ge 0$$. But there is no way to minimize $$f(x)$$ as we can keep increasing the value of $$x$$ which will make it smaller and smaller.

To show the max of $$f$$ is $$f(0)=1$$, here's one way:

$$f(x)=\frac{1}{1+x^2}\leq\frac{1}{1}=1$$ so $$1$$ is an upper bound. Therefore, $$\sup f(x) \leq 1$$. At the same rate $$f(0)=1$$ shows all upper bounds most be at least $$1$$. Therefore, $$\sup f(x)\geq 1$$. Therefore, $$\sup f(x)=1$$ and $$f(0)=\sup f(x)$$ so $$f$$ has a maximum.

For the next part, $$f(x)\geq 0$$ so $$0$$ is a lower bound. Therefore, $$\inf f(x)\geq 0$$. Suppose $$\inf f(x)>0$$. Choose $$\epsilon$$ such that $$0<\epsilon<\inf f(x)$$. Then $$\epsilon$$ is a lower bound for $$f$$ so $$f(x)\geq \epsilon$$ for all $$x$$. We arrive at a contradiction by finding $$x$$ such that $$f(x)<\epsilon$$.

Solving $$\frac{1}{1+x^2}<\epsilon \implies 1<\epsilon(1+x^2)\implies \frac{1-\epsilon}{\epsilon}\sqrt{\frac{1-\epsilon}{\epsilon}}$$ or $$x< -\sqrt{\frac{1-\epsilon}{\epsilon}}$$. Pick any such $$x$$ and we have a contradiction. Therefore, $$\inf f(x)=0$$.

If $$f$$ has a minimum, there must exist $$x_0$$ such that $$f(x_0)=0$$. But then $$\frac{1}{1+x_0^2}=0$$, which implies $$1=0$$, a contradiction. Thus, $$f$$ has no minimum.

Remark: The two statements you provided aren't enough. The space is not compact so you really have to perform the analysis.

• Thank you. study a section and then go to solve the excecises of that section. but it turns out the answer has nothing to do with that section! thank you for the clear explanation. – BesMath Mar 28 '20 at 0:12