Is $T : F \to \textsf{C}([a,b])$ defined by $Tf = f''$ continuous?

Question

$\textsf{C}([a,b])$ is a normed space with $$\| f \| := \sup \{ |f(x)| :\, a \leq x \leq b \}.$$ Let $$F := \{ f \in \textsf{C}([a,b]) :\, f'' \textrm{ is continuous and } f(a)=f(b)=0 \}.$$ Is $T : F \to \textsf{C}([a,b])$ defined by $Tf = f''$ continuous?

Answer 1

Let $f_n(x) = \sin ( { 2 \pi n x \over b-a})$, then $\|f_n\| = 1$, but $\|f_n''\| = ({ 2 \pi n x \over b-a})^2$.

Answer 2

You didn't tell the topology or norm on $F$. If it is $|f|_F=|f|_\sup+|f'|_\sup$, which is correct in the sense that it makes $F$ complete, then, yes, the differentiation map from $F$ to $C[a,b]$ is continuous in the respective topologies.

If, instead, you use the $C[a,b]$ sup-norm topology on $F$, then the differentiation map $F\to C[a,b]$ is not continuous.