Example of a linear operator whose graph is not closed but it takes a closed set to a closed set

I want an example of a linear operator $T:X\to Y$, where $X$ and $Y$ are normed linear spaces, such that graph of $T$ is not closed but $T$ maps closed sets of $X$ to closed sets in $Y$.

For functions other than linear operators it is not that difficult. If we consider $f:\mathbb R\to \mathbb R$ by $f(0)=0$ and $f(x)=1$ for all $x\in \mathbb R\setminus \{0\}$, then $f$ maps every subset of $\mathbb R$ to a closed set in $\mathbb R$ but grapf of $f$ is not closed.

Any help will be appreciated.

We know that there always exists a discontinuous linear functional $f:X\to \mathbb R$, where $X$ is an infinite dimensional normed linear space. If $f:c_0\to \mathbb R$ is a discontinuous linear functional then for every $n>0$, there exists $x^{(n)}\in c_0$ such that $|f(x^{(n)})|>n\|x^{(n)}\|_{\infty}$. Now we define $T:c_0\to c_0$ by $T(x)=(f(x),x_1,x_2,\ldots)$ for all $x\in c_0$. Then $\|T(x^{(n)})\|_{\infty}=\sup\{|f(x^{(n)})|,|x_1|,|x_2|,\ldots\}>n\|x^{(n)}\|_{\infty}$. Thus $T$ is a discontinuous linear operator. Since $c_0$ is a Banach space, therefore graph of $T$ is not closed. Is the function $T$ maps every closed set in $c_0$ to a closed subset of $c_0$?

• Start by looking for one example of a linear operator whose graph is not closed. – uniquesolution Mar 27 '17 at 9:53
• I am not getting a suitable example – Anupam Mar 27 '17 at 10:12
• so how do you imagine you will solve the problem? It seems to me that if you are looking for an object of type X with property Y, you certainly need to find first an object of type X. – uniquesolution Mar 27 '17 at 10:27
• Yes. I want an operator that is not closed. If I consider Banach spaces, then unbounded operator will do. But I am not finding such an operator – Anupam Mar 27 '17 at 10:37
• I have found such an example. But I could not establish that my function is closed. Any suggestion will be appreciated. – Anupam Mar 28 '17 at 14:15

Indeed, suppose the graph of $T$ is not closed. Then there exists a sequence $(x_n)$ that converges to $0$ in $X$ and such that $Tx_n\to y\ne 0$ in $Y$. Furthermore, we can arrange that $Tx_n\ne y$ for all $n$. Indeed, if $Tx_n$ happens to be equal to $y$, replace $x_n$ with $(1+1/n)x_n$ which does not change the limit of the sequence $(Tx_n)$.
Let $A= \{x_n:n\in\mathbb N\} \cup \{0\}$; this is a closed subset of $X$. Its image $T(A)$ has $y$ as its limit point, but does not contain $y$. Thus, $T(A)$ is not closed.