Show that the sequence of functions $(x_n)_{n≥1}$ in $C[0, 1]$ given by $x_n(t) = t^{2n} − t^{3n} , ∀t ∈ [0, 1]$ is bounded That is $C[0,1]$ equipped with the supremum metric. 
I have proven, using derivatives, that each function $x_n$ has a local maximum and local minimum at $(2/3)^{1/n}$ and $0$ respectively. I know graphically that these are the upper and lower bounds of the $x_n$'s, making $\max(d_{∞})= 0.148$ recurring, but I don't know how to show these are the global min and max over $[0,1]$.
There is also a second question, asking to prove that there is no convergent subsequence, and I have no ideas for that.
 A: We have
$$||x_n||_\infty\le \sup_{t\in[0,1]}|t^{2n}|+\sup_{t\in[0,1]}|t^{3n}|=2$$
hence $(x_n(t))$ is bounded on $[0,1]$.
A: The easier way is to note that $x_n(t)$ is bounded between $-1$ and $1$. Therefore $(x_n)$ is bounded in the sup norm.
If $x_n$ converges uniformly to some function $f$ then it is easy to see that $f$ must be zero, by taking the limit pointwise. Thus it suffices to look at the maximum of $|x_n|$ on $[0,1]$. 
You can find that the maximum is taken in $M_n=(2/3)^{1/n}$ and it is equal to $x_n(M_n) = (2/3)^2-(2/3)^3=M$. Note that the maximum does not depend on $n$, so $\|x_n -f\|_\infty = M>0 $ which means that $f_n$ does not converge uniformly to $0$. The same argument works, to prove that no subsequence converges uniformly.
This shows that both hypotheses in the Ascoli Arzela theorem are necessary. Your sequence $(x_n)$ is equibounded, but is not equi continuous, since $x_n'(t)=2nt^{2n-1}-3nt^{3n-1}$ which is close to $-n$ if $t$ is close to $1$.
A: We also have
$x_n(t) = t^{2n} - t^{3n} = t^{2n}(1 - t^n), \tag{1}$
whence, for all $t \in [0, 1]$,
$\vert x_n(t) \vert = \vert t^{2n} \vert \vert 1 - t^n \vert; \tag{2}$
taking suprema over $[0, 1]$ yields
$\Vert x_n(t) \Vert_\infty \le \Vert t^{2n} \Vert_\infty \Vert 1 - t^n \Vert_\infty, \tag{3}$
and since
$\Vert t^{2n} \Vert_\infty = 1 \tag{4}$
and
$\Vert 1 - t^n \Vert_\infty = 1, \tag{5}$
it follows from (3), (4), (5) that, for all $n$,
$\Vert x_n(t) \Vert_\infty \le 1 \cdot 1 = 1; \tag{6}$
the sequence $x_n(t)$ is thus contained in the $\Vert \cdot \Vert_\infty$ unit ball in $C[0, 1]$; that is, it is bounded.  QED.
As for the second question, I must cogitate further to respond.
Hope this helps.  Cheers,
and as ever,
Fiat Lux!!!
