Is the linear functional that sums the coefficients of $p$ continuous on $P([a,b])$ with $\|\cdot\|_\infty$. Is the linear functional that sums the coefficients of $p$ continuous on $P([a,b])$ with $\|\cdot\|_\infty$.
My attempt:
Claim: It is continuous. 
When $1\in [a,b]$:
Since a linear functional is continuous if and only if its kernel is closed, it suffices to check that the kernel subspace 
$$\bigg\{ p \in P([a,b]) : p(x) = \sum a_n x^n \text{ with } \sum a_n = 0 \bigg\} = \big\{ p \in P([a,b]) : p(1) = 0 \big\}$$
is closed.
And the result is clear if $p_n \rightarrow p$ in $\|\cdot\|_\infty$, then pointwise convergence holds which means 
$$p(1) = \lim_n p_n(1) = \lim_n 0 = 0$$
thus if $p_n$ is in the kernel, so is $p$, the kernel is closed.
For general case, can I expand the interval $[a,b]$ to $[a',b']$ such that both $[a,b]$ and $\{1\}$ are contained in $[a',b']$? Can we say $P([a,b])$ is a linear subspace of $P([a',b'])$ or is it only a subspace? If the linear functional is continuous on $P([a',b'])$, is it also continuous on $P([a,b])$? 
 A: As you have shown, the functional (let me call it $\varphi$) is continuous (it coincides with $\phi(p) = p(1)$) as long as $1 \in [a,b]$.
The coincidence with $\phi(p) = p(1)$ also allows you to conclude that the functional is not continuous as soon as $1 \notin [a,b]$.
To see this, consider the compact(!) set
$$
K := \{1\} \cup [a,b].
$$
Then $f : K \to \Bbb{R}$ defined by $f(1) = 1$ and $f|_{[a,b]} \equiv 0$ is continuous. By a theorem of (Stone)-Weierstraß, there is a sequence of polynomials $(p_n)_n$ with $p_n \to f$ uniformly on $K$ (if you know the theorem only on intervals, choose any compact interval $I \supset K$ and extend $f$ continously to $I$ (how?).)
But this implies $\phi(p_n) = p_n(1) \to f(1) = 1$ although
$$
\Vert p_n \Vert_{\infty, [a,b]} \to \Vert f \Vert_{\infty, [a,b]} = 0,
$$
because of uniform convergence.
The idea here was that the functional $\phi$ would provide a continous way of "evaluation of a function at $1$". I then used this intuition to provide a counterexample (because intuitively, the sup-norm on $[a,b]$ can not control $|f(1)|$, the argument above makes this rigorous.)
