Consider $Y=l^1$ and $X=\{(x_n) \in Y: \sum n|x_n|<\infty\}$ and the linear operator $T:X\to Y$ by $(Tx)_n=nx_n$.

I need to prove that the graph of $T$ is closed but $T$ is not continuous. The graph of $T$ is defined as $G(T)=\{(x,Tx):x\in X\}$.

How do I show that?

Furthermore the Closed Graph Theorem says " If $X$ and $Y$ are Banach spaces and $T$ a linear transformation such that $G(T)$ is closed, then T is continuous." Why can't I apply this theorem here? Is X not a Banach Space?

Also I need to show that there exists an inverse $S$ of $T$ and that $S$ is a bounded isomorphism but not boundly invertible.

Am I totally wrong if I just say that $S$ defined as $(Sx)_n=\frac x n$ is the inverse of $T$?

Again I can't apply the Bounded Inverse Theorem, which says that if $X$ and $Y$ are B.S. and T is bijective then T is invertible. Why is that?

I am happy about help!


2 Answers 2

  1. T is unbounded, because the sequence $(0,0,...,0,1,0,...)$ with a one in k`th position is mapped to k times itself. And for linear operators bounded is equivalent to continuous.

  2. X is indeed not a Banach space (it is not complete). Therefore CGT is not applicable.

  3. The inverse is $(Sx)_n=x_n/n$. It should be easy enough to show that $||S||\le1$ which implies that S is continuous .

  4. To show that the graph of an operator $T$ is closed you take a convergent sequence in X (which is a "sequence of sequences", don't get confused there), i.e. $x^k\to x\in X$ and also assume that the images under T converge, i.e. $Tx^k\to y\in Y$. You than have to show that the limits agree, i.e. $y=Tx$. You don't have to show convergence itself (that is the reason why closedness is easier than continuity).

EDIT: Here is a quick way to show closedness: $S:Y\to X$ is bounded, therefore $S$ is continuous, therefore $G(S)$ is closed. Now $G(T)$ is simply the transposed of $G(S)$, so it is closed as well.

Or, alternatively, more along the lines I explained earlier: Let $x^k\to x$ and $Tx^k\to y\in Y$. Then by continuity of $S$ we have $x^k = STx^k \to Sy$. by convergence of $x^k$, it must be $Sy=x$, i.e. $y=TSy=Tx$.

Note: I wrote the indices up in order to avoid confusion with the index inside the sequences. I.e. $x^k_n$ would be the $n$'th number inside the $k$'th sequence.

  • $\begingroup$ Thank you! I have got a question to 3. Why is S the inverse? To me it seems that T(S(x))= x and not 1? $\endgroup$
    – Yuhe
    Commented Nov 21, 2016 at 11:43
  • $\begingroup$ yep, T(S(X))=x for all $x\in X$. This is what "invertible" means. The concatenation of $T\circ S$ is equal to the identity-operator, which is sometimes written as "$1$". $\endgroup$
    – Simon
    Commented Nov 21, 2016 at 11:45
  • $\begingroup$ Oh god yeah, you are right! Could you write out a bit more to why the graph is closed? $\endgroup$
    – Yuhe
    Commented Nov 21, 2016 at 11:48
  • $\begingroup$ did some editing. $\endgroup$
    – Simon
    Commented Nov 21, 2016 at 12:09
  • $\begingroup$ Thank you a lot!! Could you maybe also help me with why S is not boundly invertible? $\endgroup$
    – Yuhe
    Commented Nov 22, 2016 at 10:23

Firstly, $T$ is not continuous because it is not bounded (recall that a linear transformation is continuous if and only if it is bounded, i.e. there is some real $c \ge 0$ such that $||T(x)||_y \le c||x||_x$ ). Take a set $S = \{s_i\} $ of sequences in $X$ of the form $s_i = (0, 0, ...1, ...)$ where the $1$ is in the $i^{th }$ position. Then $||s_i||_x = 1$ and $T(s_i) = i$ So, $||T(s_i)||_y = i = i||s_i||_x$ which is $< \infty$ for all $s_i$ (confirming that $s_i \in X$) but is not bounded.

Secondly, $X$ is not Banach because it is not complete. Consider a set of sequences in $X$ of the form $S = \{s_i\}_{i = 1, \infty} $ where $s_i = (1/n^{2 + 1/i})_{n = 1, \infty}$ .Then each $s_i \in X$ and the set of sequences converges to $s = (1/n^2)_{n = 1, \infty} $, but $s \not \in X$ because $T(x) = (n/n^2)_{i = 1, \infty} = (1/n)_{n = 1, \infty}$ which is infinite.

Thirdly, $T$ has a closed graph: to show this take set of sequences $S = \{s_i = (s_{i,n})_{n = 1, \infty} \}_{i = 1, \infty} $ which converge to a sequence $s = (s_n)_{n = 1, \infty} $: to clarify the notation, $s_{i, n}$ is the $n^{th}$ term in the $i^{th}$ sequence. Show that if $T(s_i)$ converges to some $y \in Y$ then $y = T(s)$ and $s \in X$. This is one definition of a closed graph.

$||T(s_i) - T(s) || = ||s_{i, 1} - s_1||_x + 2 ||s_{i,2} - s_2||_x + 3||s_{i, 3} - s_3||_x....$


  • $\begingroup$ Thanks a lot already!! I thought that G(T) is closed if all the limit points are also in G(T).. Can you explain your proof a bit more? $\endgroup$
    – Yuhe
    Commented Nov 21, 2016 at 11:04
  • $\begingroup$ Why is T not bounded, to me it seems that you proved that ||Tx|| = i ||x||, so bounded? Where is my knot in my head? :( $\endgroup$
    – Yuhe
    Commented Nov 21, 2016 at 11:35
  • $\begingroup$ $T$ is not bounded because there is always a larger $i$. The closed graph part is actually giving me some difficulty to complete. I may need to come back to it later. $\endgroup$ Commented Nov 21, 2016 at 11:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .