\textbf{Exercise:} Let $f : [1,+\infty)\to\mathbb{R}$ be a continuous function such that $\lim_{x\to +\infty} f(x) = +\infty$. Prove that $f$ has a minimum value.\\
\textbf{Thoughts:}\\
1. Since $\lim_{x\to +\infty} f(x) = +\infty$, the following statement holds true:\\
$\left(\forall M > 0\right)\left(\exists N > 0\right)\left(\forall x\in [1,+\infty)\right)\left(x > N\implies f(x) > M\right).$\\
2. Let $M_1 > 0$. Then, there exists some $N_1 > 0$, such that for all $x\in [1,+\infty)$, if $x > N_1$, then $f(x) > M_1$.
#Minimum Value – Possible IVT Problem
mellow cypress
elfin fieldBOT
```tex
\textbf{Exercise:} Let $f : [1,+\infty)\to\mathbb{R}$ be a continuous fun...
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hollow shoreBOT
Évariste
wide anchor
- The range of f is bounded from below. Hence it has infimum.
- f is continuous, show there exists a point that attains infimum
you might recall a continuous function in a closed interval attains its extremal values
@mellow cypress
your current argument does not appeal to continuity
continuity is required for this claim to hold
mellow cypress
Why is the range of f bounded from below?
wide anchor
you have implicitly written it in your argument already
wide anchor
you might recall a continuous function in a closed interval attains its extremal...
but you haven't mentioned this part
mellow cypress
you have implicitly written it in your argument already
I understand. So, M_1 is a lower bound of f.
wide anchor
not necessarily
you have that f(x)>M1 for all x > N1
but f is continuous in the closed interval [1, N1]
mellow cypress
you have that f(x)>M1 for all x > N1
So it is a lower bound for all x > N_1?
wide anchor
therefore..?
mellow cypress
therefore..?
By the EVT, f attains minimum and maximum.
wide anchor
correct
mellow cypress
I still do not understand what we have exactly done.
wide anchor
you look at two intervals
mellow cypress
I am very new to this, this is my first semester at college.
wide anchor
[1, N1]
(N1, inf)
mellow cypress
How do we know that N_1 not less than 1 exists? (so the interval [1,N_1] exists.)
wide anchor
you should be more precise here then if it confuses you
your function's domain is [1, inf)
mellow cypress
your function's domain is [1, inf)
I understand that.
wide anchor
so fix f(1) = a
and pick M > |a|
the N for which f(x)>M for all x>N has to be bigger than 1
mellow cypress
the N for which f(x)>M for all x>N has to be bigger than 1
Let me think a bit about that.
I do not understand.
Why does N have to be bigger than 1?
wide anchor
let's say f(1) = -1
now pick M=2
find N that guarantees f(x)>2 for all x>N
such N exists by assumption
and N=1 is impossible
but even if N=1 is possible, you can always pick something bigger
it doesn't matter
mellow cypress
It seems a bit tricky to understand.
I've been watching your thoughts as carefully as I can.
mellow cypress
Yes.(666 > 1)
wide anchor
so does it matter if N=1?
mellow cypress
What do you mean if it matters?
wide anchor
- Pick any M>0.
- By assumption there exists N>=1 such that f(x)>M for all x>N.
- If necessary, increase N so it's bigger than 1
- the function f is continuous in [1,N] therefore it attains its minimum.
so now you have broken the function in two parts: [1,N] and (N,inf)
you know there is minimum in [1,N], can you show that it's minimum in [1,inf)
mellow cypress
you know there is minimum in [1,N], can you show that it's minimum in [1,inf)
Before that, let me see if I understand all steps.
I do not understand steps 2 and 3.
wide anchor
step 2
wide anchor
Yes.(666 > 1)
step 3
wide anchor
you should be more precise here then if it confuses you
I did say you might want to be more precise if you get confused by your own writing
mellow cypress
For all positve M there exists some positive N etc., but N might always be less than 1.
wide anchor
so?
mellow cypress
That would be a problem.
wide anchor
your function's domain is [1, inf)
.
mellow cypress
Of course, but how is that linked to N being not less than 1?
wide anchor
why does it matter?
mellow cypress
Because of step 2.
mellow cypress
Yes.
mellow cypress
I understood!
wide anchor
good 🙂
mellow cypress
Now what's left is to show that the minimum of f at [1,N] is also minimum at (N,inf), as you have said above.
Question: Why not include N in (N,inf)?
wide anchor
Now what's left is to show that the minimum of f at [1,N] is also minimum at (N,...
I didn't say it exactly like that
wide anchor
Question: Why not include N in (N,inf)?
it's already in [1,N]
mellow cypress
it's already in [1,N]
I see.
mellow cypress
I didn't say it exactly like that
"you know there is minimum in [1,N], can you show that it's minimum in [1,inf)"
wide anchor
$$ (\exists a\in [1,N])(f(a) = \inf_{x\in [1,N]}f(x)) $$
mellow cypress
Using $$ swaps the background and font colours?
hollow shoreBOT
aL
wide anchor
no, I have changed my background color
mellow cypress
How?
wide anchor
beats me, I don't remember
mellow cypress
Oh, ok.
wide anchor
some server commands, you can ask in general chat
wide anchor
that means there is an element a in [1,N] which attains the minimum
mellow cypress
but it says inf.
wide anchor
yes
mellow cypress
and since inf is in f, it's min.
wide anchor
if minimum exists, then minimum = infimum
mellow cypress
I understand.
mellow cypress
if minimum exists, then minimum = infimum
Yes. What about what I said?
wide anchor
and since inf is in f, it's min.
i don't know what this means 
mellow cypress
Never mind about that. I understand what you said.
So, isn't the last step to show that the minimum of f at [1,N] is also minimum at (N,inf)?
wide anchor
that would be sufficient
but you have to be careful
I have only used continuity in [1,N] up to now
in your proof you have to use continuity everywhere
if you don't then you'll have this problem
mellow cypress
in your proof you have to use continuity everywhere
Since f is continuous everywhere(given in the problem statement), then what's the problem?
wide anchor
yes, it's continuous everywhere but we haven't explicitly used this fact
we have only said that f is continuous in some finite closed interval
mellow cypress
So, what do we do. It seems impossible for us to be able to "extend the minimum".
wide anchor
by definition of minimum, f(a) =< f(x) for all x in [1,inf), yes?
mellow cypress
Yes. That's what we want to show, right?
wide anchor
this is just an example of what happens if you don't use continuity in your proof
this is not a graph of the original situation
mellow cypress
Of course. But, if M is from the definition of the infinite limit of f at infinity, isn't it true that f(a) > M?
mellow cypress
I understand that we have shown that f(x) >= f(a) for all x in [1,N].
wide anchor
yes
mellow cypress
I am not sure how to proceed.
wide anchor
and that is the main part of this proof
mellow cypress
What's left to show is that f(x) >= f(a) for all x in (N, +inf).
wide anchor
yes
mellow cypress
I am trying to think of ideas.
For x > N, f(x) > M. Choose M = a, so f(x) >= a. for all x in [1,+inf).
wide anchor
since f is bounded from below, it has infimum, and because of continuity, f attains all values between infimum and +inf
think about M itself for example
mellow cypress
I think I completed the proof?
wide anchor
For x > N, f(x) > M. Choose M = a, so f(x) >= a. for all x in [1,+inf).
we don't know this
wide anchor
I understand that we have shown that f(x) >= f(a) for all x in [1,N].
we know this
mellow cypress
For x > N, f(x) > M. Choose M = a, so f(x) >= a. for all x in [1,+inf).
What's wrong with this?
wide anchor
how did we find a?
we first picked M, from which we found N and then found a
a is dependent on the choice of M
this is nonsense
mellow cypress
I see.
mellow cypress
since f is bounded from below, it has infimum, and because of continuity, f atta...
Seems intuitive, but what theorem guarantees that f attains all values between infimum and +inf, when it is continuous?
mellow cypress
Probably it's an extended version of I.V.T.
wide anchor
the continuous image of an interval is also an interval
mellow cypress
the continuous image of an interval is also an interval
I understand that this follows from IVT.
wide anchor
then assuming M is big enough, M itself is a value of f, yes?
mellow cypress
yes
mellow cypress
yes
wide anchor
if f(a) <= M, then we're done
because f(a) is minimum in [1,N] and f(a) <= M < f(x) for all x > N
yes?
mellow cypress
yes
wide anchor
what if f(a)>M ?
mellow cypress
Then, there is a problem.
wide anchor
why
mellow cypress
We need it to be less than or equal to M.
We have to show this leads to a contradiction.
wide anchor
i said that would be sufficient
wide anchor
We have to show this leads to a contradiction.
good luck finding one
want me to draw you a picture?
mellow cypress
sure
wide anchor
mellow cypress
Thanks, I see.
wide anchor
but we are almost done
mellow cypress
We are?
wide anchor
if f(a) <= M, then we're done
you understand this part, yes?
mellow cypress
yes
mellow cypress
yes
wide anchor
M is a lower bound for the values of f in the interval (N,inf)
mellow cypress
yes
wide anchor
therefore the values of f in (N,inf) have infimum
wide anchor
now does that imply the values of f in [N, inf) have infimum?
mellow cypress
[N,inf) or [1,inf)? I think for [N,inf) it's obvious.
wide anchor
- By continuity of f, show that f(N) >= M
- Show that there exists N' such that the infimum of f is attained within [N,N']
mellow cypress
Thanks for the help. I have to go to sleep right now(It's night here in Greece). Will try to complete the solution tomorrow. WIll ask more questions if neeeded.
wide anchor
sound good
signal voidBOT
@mellow cypress