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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...

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2 Answers 2

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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 $\{ T^{-1} x_n \}$?

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  • $\begingroup$ is the method used in post correct or not?? $\endgroup$ Commented Jan 25, 2016 at 17:53
  • $\begingroup$ 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. $\endgroup$
    – Joel
    Commented Jan 25, 2016 at 17:56
  • $\begingroup$ 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? $\endgroup$ Commented Jan 25, 2016 at 18:27
  • $\begingroup$ 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. $\endgroup$
    – Joel
    Commented Jan 25, 2016 at 19:00
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    $\begingroup$ In addition, you also said that $T^{-1}$ is continuous by similar means, but again gave no argument. You need to provide more details. $\endgroup$
    – Joel
    Commented Jan 25, 2016 at 19:01
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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 $||T^{-1}||$.

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  • $\begingroup$ :is the method used in my post wrong??? $\endgroup$ Commented Jan 25, 2016 at 17:51
  • $\begingroup$ Yes. You write "being linear it will be continuous" and this is false. $\endgroup$
    – levap
    Commented Jan 25, 2016 at 17:54
  • $\begingroup$ 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? $\endgroup$ Commented Jan 25, 2016 at 18:25
  • $\begingroup$ @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. $\endgroup$ Commented Jan 25, 2016 at 19:16
  • $\begingroup$ @vikashtripathi it's not clear how you can say "just by looking" that the operator is continuous at $(1,1,0,\dots)$ $\endgroup$ Commented Jan 25, 2016 at 19:19

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