Here are the statements:
- Suppose that $f$ is continuous on the interval $[a,b)$ and that $\lim_{x\to b^{-}} f(x) = +\infty$. Then $f(a) \le f(x)$ for all $x$ in $[a,b)$.
- Suppose that $f$ is continuous on the interval $[a,b)$ and that $\lim_{x\to b^{-}} f(x) = +\infty$. Then there exists a point $c\in [a,b)$ such that $f(c)\le f(x)$ for all $x$ in $[a,b)$.
- Suppose that $f$ is defined on the interval $[a,b)$ and that $\lim_{x\to b^{-}} f(x) = +\infty$. Then there exists a point $c\in [a,b)$ such that $f(c)\le f(x)$ for all $x$ in $[a,b)$.
Here are my arguments and proofs:
For $(1)$, we consider the the function $f\colon [-\pi /4 , \pi /2 )$ given by $f(x)=\left| \tan x \right| $. Clearly $\lim_{x \to {\pi / 2}^{-}} f(x) = +\infty$ and $ f(0) = 0 < 1 = f(- \pi / 4)$. This disproves the statement.
For $(2)$, we just need to show that the set $W=\{ f(x) \colon a\le x <b \}$ has an infimum. Then this proof follows like the proof of EVT (where $f$ is continuous on $[a,b]$). Clearly, $f(a) \in W$ so it is non empty. Now, we need to show that $W$ is bounded below. So, let $m\in \mathbb{R}$ with $m$ not being in the image of $f$. We claim that $m$ is an lower bound of $W$. Suppose not, then there is a point $x_1 \in [a,b)$ such that $f(x_1)<m$. Also since, $\lim_{x\to b^{-}} f(x) = +\infty$, there is also a point $x_2$ such that $m < f(x_2)$. Thus, by the intermediate value theorem, there is a point $c$ between $x_1$ and $x_2$ such that $f(c)=m$ which is a contradiction. Hence it must be that $m$ is a lower bound of $W$.
For $(3)$, we consider the the function $f\colon [-\pi /2 , \pi /2 )$ given by $f(x)=\tan x $ whenever $x\in (-\pi /2 , \pi /2 )$ and $f(x)=0 $ when $x=-\pi /2 $. Clearly there is no minimum for this function. Hence, this disproves $(3)$.
Can anyone point me if I went wrong somewhere (especially in $(2)$)?
In part $(2)$, you have shown that it is bounded below. Not sure if you have shown that a minimum is attained.
Since $\lim_{x \to b^-}f(x) = \infty$, we know that there is a value $k \in (a,b)$, such that $\forall x \in (k,b), f(x) \ge f(a).$
Now, we consider the minimization over $[a,k]$, then we know that there is a $c$ such that $f(c) \le f(x), \forall x \in [a,k]$ since $f$ is continuous and $[a,k]$ is compact.
In particular, $f(c) \le f(a)$.
Hence, we have found a $c$ such that $f(c) \le f(x), \forall x \in [a,b)$.