# Linear Operator in a Banach Space

I'm trying to get my head around linear operators and their usage with Banach spaces. Could someone help me understand how some properties relate to the following operator?

We have Banach space $$L^2$$, whose elements are real valued sequences $$x = (ξ_{j} ) = (ξ_{1}, ξ_{2}, . . .)$$ such that $$\sum_{j=1}^∞ |ξ_{j}|^2 < ∞$$,

Define the operator $$T : L^2 → L^2$$ as: $$(T x) = (ξj/j)$$ , for every $$x = (ξ_{j} ) ∈ L^2$$

That is, $$T x = (ξ_{1},ξ_{2}/2, ξ_{3}/3, . . . ,)$$ for every $$x = (ξ_{1}, ξ_{2}, ξ_{3}, . . .) ∈ L^2$$

It is easy to show that the operator T is linear (as it satisfies the definition). I think that the inverse of $$T$$ is $$T^{-1} x = (j ξ_{j})$$, how can I show that T is both injective and surjective?

Furthermore, how would one go about showing that T is bounded, continuous and then show what $$||T||$$ is?

• For boundedness note that $\|T(x)\|\leq \|x\|$, since each component is getting divided by a number not smaller than $1$. Note that this bound is tight at the vector $(1,0,0,...)$. Injectivity is direct, isn't it? For surjectivity look at the solutions of $T(x)=(1,1/2,1/3,...)$ – conditionalMethod Nov 11 at 20:00
• I think writing $\xi_j$ (in latex \xi_j) instead of $\xi j$ will make your post more readable. – Viktor Glombik Nov 11 at 20:08
• Injectivity is clear, just ask when is $T=0$. For surjectivity, note that your inverse $T^{-1}$ is not well defined as an automorphism of $L^{2}$, is $T^{-1}(1,1,1,...)$ well defined? This problem is precisely the reason $T$ can't be surjective - if it were it would have an inverse, and the inverse would have to have the formula you wrote. This leads to the same example in above comment. – Benny Zack Nov 11 at 20:09