Closed graph theorem seems to state that a closed operator has to be bounded?

Question

By the closed graph theorem an operator $T$ is continuous (equivalently bounded) if and only if it its graph is closed. An operator with a closed graph is called a closed operator.

So we have

$$ T \ \text{bounded} \Longleftrightarrow T \ \text{continuous} \Longleftrightarrow T \ \text{has closed graph} \Longleftrightarrow T \ \text{closed}. \quad (*) $$

But I often see closed operators mentioned in the context of unbounded operators. That is unbounded operators can be closed. But in $(*)$ above it seems that a closed operator is equivalent to a bounded operator?

Answer 1

Let $T : X\rightarrow Y$ be a linear operator from a Banach space $X$ to a Banach space $Y$. Then the closed graph theorem states that the following are equivalent

• $T$ is continuous;

• $T$ is bounded;

• $T$ is closed.

A linear operator $T : \mathcal{D}(T)\subsetneq X\rightarrow Y$ on a linear, dense domain $\mathcal{D}(T)$ does not satisfy the hypotheses of the closed graph theorem. So $T$ can be closed without being continuous. An example is the differentiation operator $T : C^1[0,1]\subset C[0,1]\rightarrow C[0,1]$. This operator is linear, densely-defined and closed, but is not bounded.