Linear Operator bounded on a basis

Question

Given a Hilbert space $\mathcal H$, a basis $\{e_j\}$ and an injective function $T$ from $\{e_j\}$ to $\mathcal H$ such that $\| T(e_j) \| \leq C$ for all $j$. Can we always extend $T$ to a bounded linear operator on $\mathcal H$?

Answer 1

No it may not work!

Procedure

The standard procedure is to first define it on a basis.

Then expanding it linearly to its span.

After checking continuity extending it onto the closure.

(As it was defined on a basis this will be well-defined and defined on the whole Hilbert space.)

Nonexample

Given the square summable sequences: $$\ell^2:=\{(x_n)_n:\sum_{n\in\mathbb{N}}|x_n|^2<\infty\}$$ Choose the canonical basis: $$\varepsilon_n:=(0,\ldots,1,0,\ldots)$$ Define an operator by: $$T\varepsilon_n:=\varepsilon_n+\varepsilon_0:\quad\|T\varepsilon_n\|\leq2$$ Then one has: $$\varphi:=\sum_{n\in\mathbb{N}}\frac{1}{n}\varepsilon_n:\quad\|T\varphi\|^2:=\|\sum_{n\in\mathbb{N}}\frac{1}{n}T\varepsilon_n\|^2=\sum_{n\in\mathbb{N}}\frac{1}{n^2}-1+\left(\sum_{n\in\mathbb{N}}\frac{1}{n}+1\right)^2=\infty$$ (Formally one has to do these latter steps separately.)