VMO

a) We have a well-known theorem: If the function $f(x)$ is continuous on the segment $[a;b]$ then $f(x)$ reaches the maximum value and minimum value at some point on that segment.

Consider the number $f(0)>0$, because ${\lim }_{x\to -\infty }f(x)={\lim }_{x\to +\infty }f(x)=0$ then there exist values $a$ small enough and $b$ large enough such that $f(x)<f(0),\forall x\le a$ and $f(x)<f(0),\forall x\ge b$.

Consider the value of $f(x)$ on $[a;b]$, by the above theorem, $f(x)$ reaches maximum value $M=f(c)$ with $M\ge f(0)$ with some $c\in [a;b]$. It is easy to see that for all $x\in (-\infty ;a)\cup (b;+\infty )$, $f(x)<f(0)\le M$ so $f(x)\le M,\forall x\in \mathbb{R}$.

Thus, $f(x)$ obtains the maximum value on $\mathbb{R}$.

b) We have the intermediate value theorem: If the function $f(x)$ is continuous and there exist two values $a<b$ such that $f(a)f(b)<0$ then $f(x)=0$ has a solution in the interval $(a;b)$.

By this theorem, it is clear that if $f(x)$ reaches two values $A,B$ for some $A<B$ then it reaches all values in the interval $(A,B)$.

Indeed, suppose that $f(u)=A,f(v)=B$ and consider a value $C\in (A;B)$ and the function $g(x)=f(x)-C$. Clearly, $$g(u)=A-C<0\text{ and }g(v)=B-C>0.$$ It follows that $g(u)g(v)<0$, so the equation has a solution in $(u,v)$.

Back to the problem, we investigate two following cases:

1) If there exists an interval $(a,b)$ containing $c$ such that $f(c)=M$ and $f(x)<M,\forall x\in (a,b)\setminus \{c\}$.

Take $A,B\ne c$ in the interval $(a,b)$ such that $c\in [A;B]$, from now on we only consider this segment. Since $f(x)$ is continuous on $[A;c]$ and $[c;B]$, there will be a minimum value on these two segments, set as $m_1,m_2$ respectively. Denote $m=\max \{m_1,m_2\}$. According to the intermediate value theorem, there exists $x_1\in [A;c]$ and $y_1\in [c;B]$ such that $f(x_1)=f(y_1)=m$; we also have $x_1<c<y_1$.

Applying this theorem again on segments $[x_1;c]$ and $[c;y_1]$, we see that there exists $x_2,y_2$ such that $$f(x_2)=f(y_2)=\frac{m+M}{2}\text{and}{x}_2<c<y_2.$$ Just like that, we consider the sequence $(u_n)$ such that $u_1=m$ and $u_{n+1}=\frac{u_n+M}{2}$ for all positive integers $n$. It is easy to see that this sequence converges to $M$ and for every $n\ge 2$, there always exists two numbers $x_n\in [x_{n-1};c],y_n\in [c;y_{n-1}]$ such that $f(x_n)=f(y_n)=u_n$ and $x_n<c<y_n$.

Note that the sequence $(x_n)$ is increasing and is bounded by $c$, so it has a limit $l\le c$. If $l<c$ then due to continuity, we have $M=\lim u_n=\lim f(x_n)=f(l)<M$, which is a contradiction. Similarly, we have $\lim x_n=c$. So two sequences $(x_n),(y_n)$ are satisfying the problem.

2) If there does not exist an interval $(a,b)$ as above, there will be a segment $[a;b]$ where $f(x)=M,\forall x\in [a;b]$.

We consider the sequence $x_n=\frac{a+b}{2}+\frac{a-b}{2^n}$ and $y_n=\frac{a+b}{2}+\frac{b-a}{2^n}$. It is easy to check that $x_n,y_n\in [a;b]$ for all positive integers $n$, so $f(x_n)=f(y_n)=M$ and $$x_n<\frac{a+b}{2}<y_n,\lim x_n=\lim y_n=\frac{a+b}{2}.$$ Thus, the problem is solved in all cases. $\square$