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Define $T: l^2 \mapsto l^2$ by $(Tx)_i = \frac{x_i}i \; \forall i=1,\ldots,n\ldots$. Prove that $T$ is a linear operator with $\|T\|=1$.

Michael Hardy
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Muneeb
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  • how do I show that T is linear? – Muneeb Mar 20 '15 at 07:35
  • You would first have to show that T(a+b)=T(a)+T(b) and kT(x)=T(kx) for something to be Linear. An operator is simply that we should remain in the same vector space (dimension). – Artem Mar 20 '15 at 07:41
  • First of all, I would suggest you rewrite the question using mathjax. It's easier to read that way. I usually just edit questions where one can figure out what you meant, but this one is slightly more confusing.

    Secondly, there are two things you need to show when proving linearity. Do you know what they are?

     – Sultan of Swing Mar 20 '15 at 07:42
  • Look at $(Tx)_i$, $(Ty)_i$ and $(T(x+y))_i$, what can you conclude? Look at $(T(kx))_i$ and $k(Tx)_i$, what can you conclude? What does the above tell you about T? For the norm part, what is norm $|Tx|$ (remember $Tx \in l^2$)? Does it remind you of some well known inequality? – TenaliRaman Mar 20 '15 at 08:14
  • Well I can conclude that T(${x_i}/{i}$)+T(${y_i}/{i}$) and T(k(${x_i}/{i}$)) = kT(${x_i}/{i}$). That is linear. – Muneeb Mar 20 '15 at 08:22
  • Am i right? @TenaliRaman – Muneeb Mar 20 '15 at 08:28
  • You seem to have the right idea, but your notation is messed up. From what is given to you, $(Tx)_i = \frac{x_i}{i}$, use it to write your earlier post clearly. – TenaliRaman Mar 20 '15 at 08:30

2 Answers2

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Linearity :

1. $T((x_i + y_i))= {(x_i+y_i)\over i}={x_i\over i}+{y_i\over i}$.

2. $T((\alpha x_i))={\alpha x_i\over i}=\alpha{x_i\over i}$

Norm :

Let $||(x_i)||=\sqrt{\sum {x_i^2 }}=1$.

$||(T(x_i))||=\sqrt{\sum {{x_i^2}\over {i^2} }}$

Observe that for each $i$, ${x_i^2\over i^2}<= x_i^2 \implies \sum {{x_i^2}\over {i^2} }<=\sum {x_i^2} = 1$.

Therefore $||T||<=1$.

Now consider the sequence $(1,0,0,...)$.

$||T((1,0,0,..))||=1$.

Therefore $||T||=1$.

Srinivas K
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Well, it's easy to see $T$ is additive, viz. if $(x_i), (y_i) \in l^2$, then

$(x_i) + (y_i) = (x_i + y_i), \tag{1}$

so that

$T((x_i) + (y_i)) = T(x_i + y_i) = (\dfrac{x_i + y_i}{i}) = (\dfrac{x_i}{i}) + (\dfrac{y_i}{i}) = T(x_i) + T(y_i); \tag{2}$

furthermore, for any scalar $\alpha$,

$\alpha(x_i) = (\alpha x_i), \tag{3}$

whence

$T(\alpha (x_i)) = T(\alpha x_i) = (\dfrac{\alpha x_i}{i}) = \alpha (\dfrac{x_i}{i}) = \alpha T(x_i); \tag{4}$

from (2) and (4) we see that

$T(\alpha (x_i) + \beta (y_i)) = T(\alpha (x_i))+ T(\beta (y_i)) = \alpha T(x_i) + \beta T(y_i); \tag{5}$

by (5), $T$ is linear. As for $\Vert T \Vert$, we have

$\Vert T(x_i) \Vert^2 = \langle T(x_i), T(x_i) \rangle = \langle (\dfrac{x_i}{i}), (\dfrac{x_i}{i}) \rangle = \sum_1^\infty \dfrac{x_i^2}{i^2} \le \sum_1^\infty x_i^2 = \Vert (x_i) \Vert^2, \tag{6}$

whence

$\Vert T(x_i) \Vert \le \Vert (x_i) \Vert \tag{7}$

for all $(x_i) \in l^2$; thus

$\Vert T \Vert \le 1. \tag{8}$

May we in fact have $\Vert T \Vert < 1$? The answer is clearly, "We may not!", since taking

$(z_i) = (\delta_{1i}) \tag{9}$

yields

$T(z_i) = (z_i) \tag{10}$

with $\Vert (z_i) \Vert = 1$,

and so

$1 = \Vert (z_i) \Vert = \Vert T(z_i) \Vert \le \Vert T \Vert \Vert (z_i) \Vert = \Vert T \Vert; \tag{11}$

it follows that

$\Vert T \Vert = 1. \tag{12}$

QED.

Note: Wrote this one up last night and fell asleep before posting; when I awoke, the fine answer of Srinivas K (and Plus One, incidentally) was already there; but I decided to post anyway, though it seems some things can only be done one way, in the hope that an even slightly different perspective may clarify further for some. Fiat Lux! End of Note.

Robert Lewis
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