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Let $\{e_k\}_{k = 1}^\infty$ be the basis of $\ell^2$. Define $T \colon \ell^2 \to \ell^2$ as follows: $$Te_k = \frac1k(e_k + e_{k + 1}),\qquad\forall k \geq 1$$ Prove that $T$ is a linear and bounded operator.

At first I thought that defining $T$ over the basis elements would be enough, but I got immediately stuck when trying to find the image of an arbitrary element $x \in \ell^2$: $$Tx = T\left(\sum_{k = 1}^\infty x_k e_k\right) = T\left(\lim_{n \to \infty} \sum_{k = 1}^n x_k e_k\right) =\ ?? \tag1$$

The solution says: $$Tx = \sum_{k = 1}^\infty x_k\frac{e_k + e_{k + 1}}k = (x_1, x_1 + \frac{x_2}2, \frac{x_2}2 + \frac{x_3}3, \ldots) \tag2$$ which is clearly linear and then proceeds to note that $T$ is the sum of two bounded operators and hence is bounded.

I really don't understand how we can go from $(1)$ to $(2)$ without requiring at least linearity and continuity, which is what we have to prove?! What am I missing?

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    Why do you try to show that $T$ is onto? – 2017-02-01
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    @NiklasHebestreit I am not. But to prove linearity, $T(\alpha x + \beta y) = \alpha Tx + \beta Ty$, we have to know how the operator behaves on an arbitrary element in $\ell^2$. – 2017-02-01
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    Yes, you are right. Sorry for that! – 2017-02-01

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$T$ is well-defined on $E= \mathrm{span} \{ e_k \}_k$, which is a dense subspace of $\ell^2$. The restriction $$T|_E :E \longrightarrow \ell^2$$ is continuous on $E$ since it is the sum of two bounded operators (on $E$).

Hence, $T|_E$ can be extended to a unique bounded operator $T$ defined on $\bar{E}= \ell^2$.

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    I see. Do you have a link to a proof of this theorem? – 2017-02-01
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    Here there is the statement https://en.wikipedia.org/wiki/Continuous_linear_extension but no proof. – 2017-02-01
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    Thanks, I found a proof [here](https://www.math.ucdavis.edu/~hunter/book/ch5.pdf) [PDF] if anyone's interested. – 2017-02-01