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let X be a real normed space with finitely many non zero terms,with supremum norm and let

T:X$\to$X be a one-one and onto linear operator defined by $$T(x_1,x_2,x_3,......)=(x_1,\frac{x_2}{4},\frac{x_3}{9},.....)$$

then, which of the following is true:

(1) T is bounded but $T^{-1}$ is not bounded

(2) T is not bounded but $T^{-1}$ is bounded

(3) both T and $T^{-1}$ are bounded

(4) neither of the two is bounded

my thought:

i am trying to apply the result :-a linear operator T:X $\to$ Y is bounded iff it is continuous.now i can say just by looking at the operator that given T is continuous at point ,say,(1,1,0,0,0,....) and being linear it will be cts on X and hence by above theorem it will be bounded.also,after constructing $T^{-1}$ and following the similar steps i am getting that $T^{-1}$ is bonded.so,(3) must be the correct option

is this approach correct??if not,please give other alternatives...

Abhishek Shrivastava
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2 Answers2

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An operator is bounded if there is a $C > 0$ st $$ \| T x \| \le C \|x\|$$ for all $x \in X$.

Now, let $e_1=(1,0,0,...), e_2=(0,1,0,0,...), $ etc. We see that $T e_n = (1/n^2) e_n$ for each $n \in \mathbb{N}$. This means $T^{-1} e_n = n^2 e_n$.

Notice that $\| T^{-1} e_n \| = n^2 \|e_n\|$. What does this say about boundedness?

Here is another approach. If $T^{-1}$ is bounded then, as you say, it is continuous. Consider the sequence $x_n = (1/n) e_n$. Notice that $\|x_n\| = 1/n \to 0$, and $x_n$ converges to zero. If $T^{-1}$ were continuous, then $T^{-1} x_n$ would converge.

What happens to the sequence $\{ T^{-1} x_n \}$?

Joel
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  • is the method used in post correct or not?? – Abhishek Shrivastava Jan 25 '16 at 17:53
  • No it is not correct. You didn't address the boundedness question appropriately. Your claim of "just looking at the operator" to see it's continuous is not sufficient. It takes more than just being linear to be continuous. For example, the differential operator is not continuous on $C^1[0,1]$, even though it is linear. The answers posted by @levap and myself outline an approach that could guide you in the right direction. – Joel Jan 25 '16 at 17:56
  • i am confused because of the following theorem: let X and Y be normed spaces over field F .and T:X→→Y be a linear operator.then,the following are equivalent:(a)T is continuous at origin(or at any point of X) (b) T is continuous on X (c)T is uniformly continuous on X (d)T is bounded .so,from this point how am i wrong? – Abhishek Shrivastava Jan 25 '16 at 18:27
  • You did not explicitly show that $T$ was continuous at the point you mentioned; you simply stated that it was obvious. Unless you write more, I cannot critique your reasoning. Demonstrating that the function is bounded is probably the best way of going about showing that $T$ is continuous. – Joel Jan 25 '16 at 19:00
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    In addition, you also said that $T^{-1}$ is continuous by similar means, but again gave no argument. You need to provide more details. – Joel Jan 25 '16 at 19:01
  • at least i got someone answering my point....thanks a lot sir.... – Abhishek Shrivastava Jan 26 '16 at 02:21
  • @vikashtripathi What both me and levap are hinting at in our posts is that $T^{-1}$ is not continuous (i.e. not bounded). Your claim "$T^{-1}$ can be seen to be continuous by similar means" tells us that the initial reasoning (that was not given) to show that $T$ was continuous must have a gap. – Joel Jan 26 '16 at 16:43
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Note that $X$ is infinite dimensional and so not all the linear operators are automatically continuous. The operator $T$ will be continuous if and only if

$$ ||T|| := \mathrm{sup}_{||x|| = 1} ||Tx|| < +\infty. $$

If $||x|| = 1$ then $|x_n| \leq 1$ for all $n \in \mathbb{N}$ and then $|(T(x_n))_m| = \left| \frac{x_n}{m^2} \right| \leq \frac{1}{m^2} \leq 1$ for all $m \in \mathbb{N}$ and thus $||T|| \leq 1$ so $T$ is continuous.

To see whether $T^{-1}$ is continuous, write $T^{-1}$ explicitly and try to estimate $||T^{-1}||$.

levap
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  • :is the method used in my post wrong??? – Abhishek Shrivastava Jan 25 '16 at 17:51
  • Yes. You write "being linear it will be continuous" and this is false. – levap Jan 25 '16 at 17:54
  • i am confused because of the following theorem: let X and Y be normed spaces over field F .and T:X$\to$Y be a linear operator.then,the following are equivalent:(a)T is continuous at origin(or at any point of X) (b) T is continuous on X (c)T is uniformly continuous on X (d)T is bounded .so,from this point how am i wrong? – Abhishek Shrivastava Jan 25 '16 at 18:25
  • @vikashtripathi not every linear operator is continuous at the origin. Not every linear operator is continuous on $X$. Not every linear operator is uniformly continuous on $X$. Not every operator is bounded. – Ben Grossmann Jan 25 '16 at 19:16
  • @vikashtripathi it's not clear how you can say "just by looking" that the operator is continuous at $(1,1,0,\dots)$ – Ben Grossmann Jan 25 '16 at 19:19
  • @ Omnomnomnom:it was really a mistake to say that...but if i show that for every sequence {$x_n$} converging to (1,1,0,0,0,0,......) ,we have T({$x_n$}) $\to$ T(x) ,then ,it will be correct to say that T is continuous from there – Abhishek Shrivastava Jan 26 '16 at 02:25