Is twice differentiation operator with given condition bounded? Let $X=\{f\in C[a,b]| f''-continuous\quad f(a)=f(b)=0\}$ and define $T:X\to C[a,b]$ by $$Tf=f''$$
I have showed that $T$ is bijective. However to show that if it is norm isomorphism or not.
How to show $||Tf||=||f||$ positively or negatively.
My work:
Think $$f(x)=[(x-a)(x-b)]^n, \quad n>2$$
$f(a)=f(b)=0$ and $f$ is clearly infinite times differentiable so $f\in X$
But if I take twice derivative of $f$
$$f''(x)=n[(x-a)(x-b)]^{n-2}(2x-(a+b))\left[(n-1)(2x-(a+b))\\+2(x-a)(x-b)\right]\\ f''(x) \sim n^2 g(x)+\mathcal O(n)$$ for some $g(x)$ bounded function on $x\in [a,b]$
so $||Tf||=||f''||\to \infty$
But I think with supremum norm $||f||_\infty$ is also not bounded. By the way without any operator I didnot consider when just a norm on a space( Banach or not) is bounded or not?
How can I show that these two norm $||Tf||$ and $||f||$ are equal or not so that $X$ and $C[a,b]$ are isometrically isomorphic or not.
 A: Depending on the norm of $X$ you get either a positive or a negative answer. That is, if you see $X$ as a subspace of $C[a,b]$ or as an entirely new space.
1. $T$ is not bounded on $(X, \|\cdot\|_\infty)$, where $\|\cdot\|_\infty$ is the usuall supremum norm:
Let $f_n(x) = \sin\bigl(n\pi\frac{x - a}{b-a} \bigr)$ for $x\in [a,b]$.
Then, we have $\|f_n\|_\infty = 1$ and
$$ \|Tf_n\|_\infty = \Bigl|\frac{n\pi}{b-a}\Bigr|^2 \|f_n\|_\infty \to \infty. $$
2. $T$ is an isometry on $(X, N)$ for $N(f) = \|f''\|_\infty$:
As $N(f) = \|Tf\|_\infty$, I only show that $N$ is a norm, in fact $(X,N)$ is complete.
As differentiation is linear, $N$ is a semi-norm. We need only prove that $N$ is definite. Let $f\in X$ with $N(f) = 0$. Then, by integration, it follows that $f$ is polynomial of degree $\le 1$. As $f$ has two distinct zeros, $f$ itself is the zero-function.
$(X, N)$ is complete: Let $f_n$ be a Cauchy sequence with respect to $N$. Then, $f_n''$ is a Cauchy sequence with respect to the supremum norm. Thus, it has a limit $h\in C[a,b]$. Integrate $h$ twice, we obtain some $g$ with $g(a) = 0$ and $g'' = h$. Now, for $f(x) = g(x) - \frac{x - a}{b - a}g(b)$, we have $f(a) = f(b) = 0$, and $f'' = h$. That is, $f$ is in $X$ and the limit of $f_n$.
