# An Example of a Linear Operator with a Closed Graph that is Unbounded

Recall from The Closed Graph Theorem that if $X$ and $Y$ are Banach spaces and if $T : X \to Y$ is a linear operator then $T$ is bounded if and only if $\mathrm{Gr}(T)$ is closed, that is, if $(x_n)$ is a sequence in $X$ that converges to $x \in X$ and $(T(x_n))$ converges to $y$ in $Y$ then $T(x) = y$.

Let $C[0, 1]$ denote the space of continuous real-valued functions on $[0, 1]$ and let $C^1[0, 1]$ denote the space of differentiable functions on $[0, 1]$. Equip both spaces with the supremum norm. Let $T : C[0, 1] \to C^1[0,1]$ be defined for all $f \in C[0, 1]$ by:

(1)It is easy to show that $T$ is linear. Let $f, g \in C[0, 1]$ and let $\alpha \in \mathbb{R}$. Then:

(2)We now show that $\mathrm{Gr}(T)$ is closed. Let $(f_n)$ be a sequence in $C[0, 1]$ that converges to $f \in C[0, 1]$ and let $(T(f_n)) = (f_n')$ converge to $g \in C^1[0,1]$. Then:

(3)The convergence here is uniform convergence, and so $\lim_{n \to \infty} f_n'(t) = g(t)$ for all $t$, we have that $f' = g$. So $T(f) = f' = g$. Hence $\mathrm{Gr}(T)$ is closed.

We will now show that $T$ is unbounded. Let $(f_n) = (x^n)$. For each $n \in \mathbb{N}$ we have that:

(4)Now observe that $(f_n') = (nx^{n-1})$. We have that:

(5)So for every $n \in \mathbb{N}$ there exists an $f_n \in C[0, 1]$ such that $\| T(f_n) \| = n$. Hence $T$ cannot be bounded! Note that this does not contradict the Closed Graph theorem since $C^1[0, 1]$ is not a Banach space!