# $T$ be the operator from $C[0,1]$ to $C[0,1]$ defined by $Tf = f'+f''$. Show that the operator $T$ is unbounded.

$f \in C[0,1]$, the space of all continuous, complex-valued functions on $[0,1]$ with supremum norm.

$\|f\|=\sup_{x\in[0,1]}|f(x)|$.

Let $D$ be the set of $f \in C[0,1]$ such that the first derivative and second derivative are defined and continuous on $[0,1]$.

Let $T$ be the operator from $C[0,1]$ to $C[0,1]$ defined by $$Tf = f'+f''$$ on the domain $D$.

Show that the operator $T$ is unbounded by constructing a sequence $f_n$ such that $\|f_n\| \le M$ for some constant $M$ but $\|Tf_n\| \rightarrow \infty$ as $n \rightarrow \infty$

• any thoughts at all? – Ittay Weiss Nov 13 '13 at 13:26
• Let $\lambda > 0$ and $f(t) = e^{\lambda (t-1)}$, then $\|f\| = 1$ but $\|Tf\| = \lambda + \lambda^2$ and you can make $\lambda + \lambda^2$ as large as you like. – achille hui Nov 13 '13 at 13:33
• You can consider $f_n(x)=x^n$. – Mhenni Benghorbal Nov 20 '13 at 22:00

Take $f_n(x)=e^{-nx}$ so $T(f_n)=-nf_n+n^2f_n=n(n-1)f_n$. Can you verify now that $T$ is unbounded?
Hint: The derivative is very sensitive to small changes, but the supremum norm is not. In particular, you can have a function that is very close to being constantly $0$ (and thus of very small norm), while the graph of it may be extremely erratic with lots of steep increasing and decreasing regions. The derivative will thus be large, and thus have large norm.