$(P[0,1],\|\|_{\infty})$ be the norm linear space Let $(P[0,1],\|\|_{\infty})$ be the norm linear space and $T$ be the differentiation operator on it. Then
$1.$ $T$ is onto right? but NOT injective as $\ker T=\{\text{ all constants }\}$ 
$2.P[0,1]$ is an Infinite dimensional Banach Space.
$3.$ $T$ is continuous on $P[0,1]$ as whenever $p_n(x)\to p(x) $ we see easily that $T(p_n(x))\to T(p(x))$, am I right?
$4.$ Is $T$ closed map? I don't know how to deal this here.
Thank you for help.
 A: *

*You are exactly correct, or equivalently you can note that any polynomials with differing constant terms are mapped to the same thing.

*Partially correct. It is infinite-dimensional. Assume you have some finite basis. Then your basis will contain a polynomial of maximum degree. But clearly we can't obtain a polynomial of strictly larger degree from our basis, so we have a contradiction and our basis will be infinite. However, our space is NOT Banach! To see this, we'll show that it's not a closed subspace of $C[0,1]$, and as a subspace of a Banach space is Banach if and only if it's closed, the result follows. Note that the sequence $f_n=1+\dots +\frac{x^n}{ n!} $ converges uniformly to $e^x$ in $C[0,1]$, but $e^x$ is not a polynomial on $[0,1]$, and thus the subspace is not closed.

*This actually isn't true. A linear operator on a normed space is continuous if and only if it is bounded. However, for arbitrarily large $n$, note that $||x^n||_{\infty}=1$, but $||Tx^n||_{\infty}=n$, so we see that $T$ is not bounded and therefore not continuous. To see where your reasoning fails, let $f_n=\frac 1 n x^n$. We have that $f_n \to 0$ uniformly. However, $Tf_n=x^{n-1}$, which does not go to $T(0)=0$ uniformly, since $T(1)=1$. In fact, $Tf_n$ doesn't converge in $P[0,1]$, and thus we see $T$ is not continuous.

*This is true. We can show an operator is closed by showing its graph $X\times T(X)$ is closed in $X \times X$. We do this as follows: let $f_n \to f \in P[0,1]$ and $T(x_n)\to y \in P[0,1]$. By the (pointwise) convergence of $f_n$ to $f$ and the uniform convergence of the derivatives to $y$, we can switch the limit and derivative to obtain $T(f)=y$. Thus we have if $(f_n, Tf_n)\to (f,y)$, then $(f,y)=(f,Tf) \in X \times T(X)$, and thus the graph is closed and $T$ is a closed map. (For the proof that we can switch the limit and derivative, see for example Rudin Ch. 7).
