I've got two questions.
(1) Define $ \|f\| := |f(0)| + \max \limits_{0 \le t \le 1} |f'(t)|$ to be a norm on $C^1([0,1])$. (I have shown it's indeed a norm).
Show that $(C^1([0,1]), \|\cdot\|)$ is complete. Here's my work:
Let $\epsilon>0$ and $(f_n)$ be cauchy in $(C([0,1]), \|\cdot\|)$. Then, for all $m,n$ greater than some $N$, we have $$ \|f_n - f_m\|<\epsilon \implies |f_n(0)-f_m(0)|<\epsilon \text{ and also } \| f_n' - f_m'\|_\infty < \epsilon. $$ By completeness of $\mathbb{R}$ and $(C([0,1]), \|\cdot\|_\infty)$, respectively, we get $f_n(0) \to a_0$ and $f_n' \to g$.
Now, we want to show $f_n \to a_0 + \int g$ or something, but I'm stuck. I'm guessing there's some ODE theory that I don't know.
(2) Let $P([0,1])$ be the set of all polynomials on $[0,1]$ with the same norm as above. Is $\phi(f) = f''(1/2)$ a bounded linear functional?
I have shown that the operators $f \mapsto f(1/2)$ and $f \mapsto f'(1/2)$ are bounded linear functionals. But I can't do it so easily with $\phi$, and I've been trying to find an unbounded sequence for a long while . . .
For Q1, by your argument, you have shown $f_n(0)\to a$ and $f_n'\to g\in C([0,1])$ in sup-norm. Then we define $f(t)=a_0+\int_0^t g(s)\,ds$ for $t\in [0,1]$. Then by fundamental theorem of calculus, it's clear that $f\in C^1([0,1])$ and $f'=g$. Then it follows that $f_n\to f$ in your norm.
Some thoughts about Q2. If we consider $f''(1)$ for simplicity, then the functional is not bounded since $f_n=\frac{1}{n+1}x^{n+1}$ has norm $1$ but $f''_n(1)=n$. I think a similar counterexample can be constructed at the point $\frac{1}{2}$.