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

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:

\begin{align} \quad T(f) = f' \end{align}

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

\begin{align} \quad T(f + g) &= [f + g]' = f' + g' = T(f) + T(g) \\ T(\alpha f) &= [\alpha f]' = \alpha f' = \alpha T(f) \end{align}

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:

\begin{align} \quad \lim_{n \to \infty} \| T(f_n) - g \| = \lim_{n \to \infty} \sup_{t \in [0, 1]} |f_n'(t) - g(t)| = \lim_{n \to \infty} \| f_n' - g \| = 0 \end{align}

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:

\begin{align} \quad \| f_n \| = \sup_{x \in [0, 1]} |x^n] = 1 \end{align}

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

\begin{align} \quad \| T(f_n) \| = \| f_n' \| = \| nx^{n-1} \| = \sup_{x \in [0, 1]} nx^{n-1} = n \end{align}

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!

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