I am considering an operator $K\colon \ell^2 \to \ell^2$ given by $$Kx = \sum_{n=1}^\infty e^{-n} \langle x , e_n\rangle e_n $$ where $e_n = (\delta_{k,n})_{k\in \mathrm{N}}$ is the standard basis on the sequence space $\ell^2$ and $ \langle \cdot , \cdot \rangle$ denotes the usual inner product. I know that this operator is bounded with $ \Vert K \Vert = e^{-1}$. Now my textbook tells me that linearity of $K$ is easily shown using boundedness, yet I am not sure whether boundedness is strictly necessary to prove this. \begin{align*} K(x+y) &= \sum_{n=1}^\infty e^{-n} \langle x + y, e_n\rangle e_n\\ &= \sum_{n=1}^\infty e^{-n} (\langle x, e_n\rangle + \langle y, e_n\rangle) e_n\\ &= \sum_{n=1}^\infty e^{-n} \langle x, e_n\rangle e_n + \sum_{n=1}^\infty e^{-n} \langle y, e_n\rangle e_n \quad (*)\\ &= Kx + Ky. \end{align*} I suspect boundedness is used in $(*)$, so making it more precise I get $$ \left\Vert \sum_{n=1}^N e^{-n} \langle x + y, e_n\rangle e_n - Kx - Ky \right\Vert \leq \left\Vert \sum_{n=N+1}^\infty e^{-n} \langle x, e_n\rangle e_n \right\Vert + \left\Vert \sum_{n=N+1}^\infty e^{-n} \langle y, e_n\rangle e_n \right\Vert$$ by the triangle inequality. Now boundedness of $K$ shows that the right hand side converges to zero as $N \to \infty$. It seems to me that boundedness is unnecessary here, and that the only thing needed is that $Kx \in \ell^2$ for all $x \in\ell^2$. Is this correct? If so, I have a follow-up question: are there examples of unbounded operators where the above argument may be used to prove linearity?

I was thinking of the defining $Tx = \sum_{n=1}^\infty n \langle x , e_n \rangle e_n$ as an example of such an operator, but I am not sure if this operator is well-defined on $\ell^2$

  • 1
    $\begingroup$ Most unbounded operators worth considering on $\ell^2$ do not have the property that $Kx \in \ell^2$ for all $x \in \ell^2$. Your operator, is not defined on $\ell^2$, but is defined on a dense domain in $\ell^2$. $\endgroup$ – Omnomnomnom Jun 14 '16 at 14:56
  • $\begingroup$ I'd say this : your second operator $(T x)_n = n \langle x,e_n \rangle$ is a densely defined unbounded operator $l^2 \to l^2$ and is well-defined as a linear operator $l^2 \to \mathbb{C}^\mathbb{N}$, but it sends some elements of $l^2$ to sequences of $\mathbb{R}^\mathbb{N}$ having an infinite $\|.\|_{l^2}$ norm $\endgroup$ – reuns Jun 14 '16 at 21:29
  • $\begingroup$ note that some densely defined unbounded operators $l^2 \to l^2$ are not even well defined as linear operators $l^2 \to \mathbb{C}^\mathbb{N}$, for example $S x = e_1 \sum_{n=1}^\infty \langle x,e_n \rangle$ $\endgroup$ – reuns Jun 14 '16 at 21:32

Boundedness gives you linearity on the entire space once you show it on a dense subspace. For example, let $\mathcal{S}$ be the subspace consisting of all finite linear combinations of the standard basis elements. Verifying linearity on $\mathcal{S}$ is straightforward because there are no convergence issues. Then, if $x,y$ are general, you can let $x_n = \sum_{k=1}^{n}\langle x,e_n\rangle e_n$ and $y_n=\sum_{k=1}^{n}\langle y,e_n\rangle e_n$ and argue from boundedness or continuity that \begin{align} K(\alpha x+\beta y)&=K(\lim_n (\alpha x_n+\beta y_n)) \\ &=\lim_n K(\alpha x_n+\beta y_n) \\ &=\lim_n (\alpha Kx_n +\beta Ky_n) \\ &=\alpha \lim_n Kx_n +\beta \lim_n Ky_n \\ &=\alpha K\lim_n x_n + \beta K\lim_n y_n \\ &=\alpha Kx + \beta Ky. \end{align}


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