Prove that $\exists c > 0$ such that $||f||_2 \le c ||T(f)||_{\infty}$ Problem (Unsolvable, since the statement is false.)
Let $X = (C^2([0,1]),||\cdot||_2)$, where $||f||_2 = ||f''||_{\infty} + ||f'||_{\infty} + ||f||_{\infty}$.
Define $T:X\to C([0,1])$ by: $\ \ \ \ T(f) = f'' + af' + bf$
where $a,b \in C([0,1])$ such that $T$ is surjective.
Prove that $\exists c > 0$ such that $||f||_2 \le c ||T(f)||_{\infty}$
What I've gathered
I know that $X$ is complete and that $T$ is linear. $T$ is bounded as well, since for $||f||_2 = 1$:
$||T(f)||_\infty = ||f'' + af' + bf||_\infty \le ||f''||_\infty + A ||f'||_\infty + B ||f||_\infty \le (1+A + B)$.
where $A = ||a||_\infty, B=||b||_\infty$. So far, I know that $T$ is a bounded linear map, so I know that:
$\forall f \in X: ||T(f)||_\infty \le ||A||\cdot||f_2||$
So what I need to prove is that $\exists c > 0$ such that $\forall f \in X:$
$\frac{||T(f)||_\infty}{||A||} \le ||f||_2 \le c ||T(f)||_\infty$
i.e. that $||\cdot||_2$ and $||T(\cdot)||_\infty$ are equivalent norms, right?
I have no idea how to proceed from here..
Edit
Since I got a counterexample showing that above is not true in general, I show below the original question (taken from an exam), which I shortened, but maybe changed the question in doing so:
Original Problem
Consider the Banach space $C^2([0,1])$ with $||\cdot||_2$ as defined above. Let $a,b \in C([0,1])$ and assume that for every $g\in C([0,1])$, the differential equation $f''+ a f' +b f = g$ has a solution $f\in C^2([0,1])$. Prove that there exists $c > 0$, such that for every $g \in C([0,1])$, the above equation has a solution $f\in C^2([0,1])$, with $||f||_2 \le c ||g||$
Problem $\neq$ Original Problem
I realize now that the property $ \forall f: ||f||_2 \le ||T(f)||_\infty $ is not explicitly required. My bad.
 A: This result is false, here is a counterexample :
Take $a=  0$ and $b=1$. We first show that $T$ is surjective.
Let $g\in C([0,1])$. Then :
$$f(t) = \int_0^t \sin(t-s)g(s)\text ds$$
is in $C^2([0,1])$ and has $Tf = g$.
Then, if you take $f = \cos$, you get $T(f) = 0$ but $f\neq 0$, so there can be no constant $c>0$ such that :
$$\|f\|_2 \leq c \|Tf\|_\infty$$
More generally, it seems that for any $a,b \in C([0,1])$, the theory of linear differential equations guarantees that $T$ has a non trivial kernel, so that this result is false in every case.
Edit : new question
To show that there is a constant $c >0$ such that for any $g\in C([0,1])$ there is a $f\in C^2([0,1])$ such that $T f =g$ and $\|f\|_2 \leq c\|g\|_\infty$, we rewrite the equation $f'' + af' + bf = g \tag{1}$ as a first order differential equation : let
$$X = \begin{pmatrix} f' \\ f\end{pmatrix}, \quad A = \begin{pmatrix} -a & -b \\ 1 & 0\end{pmatrix} \quad \text{and}\quad B = \begin{pmatrix} g \\ 0 \end{pmatrix}$$
Then $(1)$ is equivalent to :
$$X' = AX +B \tag 2$$
Let $U \in C^1([0,1],\mathcal M_2(\mathbb R))$ be then solution to the Cauchy problem :
$$\left\{ \begin{array}{ll} U'= AU \\U(0) = \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}\end{array}\right.$$
It exists (and is unique) by the Cauchy-Lipschitz theorem. Then, we can check that :
$$X(t) = U(t)\int_0^t U^{-1}(s) B(s)\text ds$$
is a solution of $(2)$. There is therefore $f\in C^2([0,1])$ such that :
$$X  = \begin{pmatrix} f' \\f\end{pmatrix}$$
Then, with $\| \cdot \|_1$ the $L^1$ norm on $\mathbb R^2$ and $|||\cdot|||$ the associated operator norm, we have :
\begin{align}
\| f \|_\infty + \|f' \|_\infty &= \sup_{t\in [0,1]} \|X(t)\|_{1} \\
&\leq C\| g\|_\infty
\end{align}
with $C = \sup_{t\in [0,1]} ||| U(t)|||  \times \sup_{t\in [0,1]}|||U^{-1}(t) ||| < \infty$.
To conclude, we compute:
$$\| f'' \|_\infty = \| g \| + \|a\|_\infty \|f'\|_\infty + \| b\|_\infty \|f\|_\infty$$
so :
$$\|f\|_2 \leq (1 + \max(1+\|a\|_\infty, 1+\|b\|_\infty)C) \|g\|_\infty$$
A: This is true provided that $T$ is injective, i.e., $Tf=0\Longrightarrow f=0$.
In such case, since $T$ is surjective, and both $C[0,1]$ and $C^2[0,1]$ are Banach spaces, then $T$ is open, due to Open Mapping Theorem, and hence $T^{-1}$ is continuous and therefore bounded, and thus there exists a $c>0$, such that
$$
\|T^{-1}y\|_2\le c\|y\|_\infty, \quad\text{for all $y\in C[0,1]$}
$$
equivalently
$$
\|x\|_2\le c\|Tx\|_\infty, \quad\text{for all $x\in C^2[0,1]$}.
$$
