Question: Consider the sequence of functions $(f\_n)\_n$ given by $$ f\_n : [0,1) \rightarrow \mathbb{R}: x \mapsto x^n. $$ I need to determine if this sequence converges pointwise and/or uniform.

First, I think it converges pointwise to the zero function, since $\lim\_{n \to \infty} f\_n = 0$ for every $x \in [0, 1)$. I could not complete my proof though.

Let $x \in [0, 1)$ be arbitrary. Let $\epsilon > 0$. Then we have to find a $n\_0 \in \mathbb{N}$ such that from $\forall n \geq n\_0$ it follows that $|f\_n(x) - 0| < \epsilon$. Now we have $$|f\_n(x) - 0| = x^n $$ Should I now consider the separate cases: $x = 0$ and $x \neq 0$? If we let $x \neq 0$ first, then we would have $x^n < \epsilon $ if $n \ln(x) < \ln(\epsilon)$, i.e. if $n > \ln(\epsilon) / \ln(x)$. Then I would choose $$ n\_0 > \ln(\epsilon) / \ln(x). $$The case $x = 0$ is trivial. 

For uniform convergence, I have no idea. But is the above approach correct?

Answer: 
Note that if $p(x)$ is degree $d\ge1$, $xp(x)+p(0)$ is degree $(d+1)$, and so cannot be equal to the $0$ polynomial. 

Furthermore, if $p$ is nontrivially degree $0$, i.e. a nonzero constant, then $xp(x)+p(0)$ is degree $1$, and again thus cannot be equal to the $0$ polynomial.

So, $p$ must be degree $0$, and cannot be equal to a nonzero constant, hence must be identically $0$.
If $L\colon P\_2(\Bbb R) \to P\_3(\Bbb R)$ and $p(x) = ax^2+bx+c$, we must solve $$L(p(x)) = ax^3+bx^2+cx+c = 0 \in P\_3(\Bbb R)$$ for $p(x)$ and not for $x$, so this must be the zero polynomial. So immediately we must have $a=b=c=0$ and $L$ is injective.

If you fix $p(x) \in P\_2(\Bbb R)$ and define $L\colon \Bbb R \to \Bbb R$ by that expression, then you must solve the third degree equation.