Closed linear operators vs Continuous linear operators Suppose we have two real Banach spaces $X, Y$, and a linear operator $A:X \rightarrow Y$. We say that $A$ is closed if whenever $u_k \rightarrow u$ in $X$ and $A u_k \rightarrow v$ in $Y$, then $Au = v$. This definition is very reminiscent of the sequential criterion for continuity of real-valued functions.
I assume that there are operators that are closed but not continuous, because otherwise, there'd be no point in having a different word. If my assumption is correct, is this because operators behave differently when we consider more general spaces, or because there is some subtle difference between this definition and the sequential criterion that I'm missing? Also, could someone provide an example of such a function?
 A: This is much later than 2017 but I thought an important question to address.
Let us relax the situation to $X,Y$ are just normed spaces to illustrate the differences.

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*As you pointed out, we say that a linear map $A:X\to Y$ is closed if whenever $x_k\to x$ in $X$ and $Ax_k \to y$ in $Y$, we have $Ax=y$ (i.e. we have $Ax_k\to Ax$ in $Y$).

*To recap, a linear map $A:X\to Y$ is continuous if whenever $x_k\to x$ in $X$, we have $Ax_k\to Ax$ in $Y$.

The subtle difference is that the definition for closed only needs us to check the sequence $\{x_k\}$ when we also know that $\{Ax_k\}$ converges to something. On the other hand, continuity forces the convergence of $\{Ax_k\}$ as part of the conclusion. From these definitions, it is easy to see that a continuous map is closed.
Now, if $X,Y$ are Banach spaces, then the two definitions are in fact equivalent, but the fact that closed implies continuous is not easy to see and is what is known as the closed graph theorem.
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Let $X,Y$ be just normed spaces. For OP's further study next time, sometimes a linear operator is only defined on a subspace of $X$, i.e. $A:D(A)\to Y$ where $D(A)\subseteq X$. Here, the definitions are more subtle:

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*A linear map $A:D(A)\to Y$ is closed if whenever $x_k\to x$ in $X$ where $x_k\in D(A)$ and $Ax_k \to y$ in $Y$, we have $x\in D(A)$ and $Ax=y$.

*A linear map $A:D(A)\to Y$ is continuous if whenever $x_k\to x$ in $D(A)$ (read again: in $D(A)$), we have $Ax_k\to Ax$ in $Y$
Without further assumptions, the two notions cannot be compared. As before, the sequences checked by continuity does not need $\{Ax_k\}$ to converge a priori. On the other hand, sequences checked by continuity needs the convergence of $x_k$ to happen in $D(A)$ (i.e. $x\in D(A)$) whereas that is part of the conclusion for sequences checked by closedness. One can find counterexamples where $A$ closed does not imply $A$ continuous and vice versa.
However, if $D(A)$ is closed, then:

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*$A$ continuous implies $A$ closed (similar reason as before, and $x\in D(A)$ is supplied by the closedness of $D(A)$

*If further $X,Y$ are Banach spaces, then $A$ closed implies $A$ continuous. This is the closed graph theorem
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Tsemo's answer above is able to give a closed but not countinuous operator because the domain of that operator does not cover all of $l^2$. For example, the operator cannot act on the sequence $(1,1/2,1/3,...)\in l^2$.
A: Hint: Consider $l^2$, the operator defined by $L(e_i)=ne_i$ is closed but not bounded.
