# Any examples of unbounded linear operators between $\ell^\infty$ and $\ell^\infty$? $\ell^p$ and $\ell^p$? $\ell^p$ and $\ell^\infty$?

Let $X$ and $Y$ be normed spaces, either both real or both complex. Let $f \colon X \to Y$ be a linear operator. Then $f$ is said to be bounded if there exists a real number $r > 0$ such that $$\lVert f(x) \rVert_Y \leq r \lVert x \rVert_X \mbox{ for every } x \in X.$$ If there is no such $r$, then $f$ is said to be unbounded.

Now my question is, can we find an example of an unbounded linear operator (i) $f \colon \ell^\infty \to \ell^\infty$? (ii) $f \colon \ell^p \to \ell^p$, where $p$ is such that $1 \leq p < +\infty$? (iii) $f \colon \ell^\infty \to \ell^p$? (iv) $f \colon \ell^p \to \ell^\infty$? (v) $f \colon \mathrm{C}[a, b] \to \mathrm{C}[a, b]$?

Or, examples of unbounded linear operators between any other pairs of these normed spaces?

By definition, $\ell^\infty$ is the normed space of all the bounded sequences of (real or complex) numbers, with the normed defined by $$\left\lVert \left( \xi_n \right)_{n \in \mathbb{N} } \right\rVert \colon= \sup \left\{ \ \left\lvert \xi_n \right\rvert \ \colon \ n \in \mathbb{N} \ \right\}.$$

For any real number $p$ such that $1 \leq p < +\infty$, the normed space $\ell^p$, by definition, is the vector space of all the sequences $\left( \xi_n \right)_{n \in \mathbb{N} }$ of (real or complex) numbers, for which the series $\sum \left\lvert \xi_n \right\rvert^p$ converges, that is, $$\sum_{n=1}^\infty \left\lvert \xi_n \right\rvert^p < +\infty,$$ with the norm given by the formula $$\left\lVert \left( \xi_n \right)_{n \in \mathbb{N} } \right\rVert \colon= \sqrt[p]{ \sum_{n=1}^\infty \left\lvert \xi_n \right\rvert^p }.$$

For any real numbers $a$ and $b$ such that $a < b$, the space $\mathrm{C}[a, b]$ is the normed space of all the real or complex-valued functions defined and continuous on the closed interval $[a, b]$ of the real line, with the norm defined by $$\lVert x \rVert \colon= \max \{ \ \lvert x(t) \rvert \ \colon \ a \leq t \leq b \ \}.$$

Right now, the only example of an unbounded linear operator that I can recall is that of the differentiation operator of the normed space of all the continuously differentiable functions on a closed interval $[a, b]$ with the maximum norm into this normed space itself.

I would appreciate references to some elementary-level text on analysis where examples of such unbounded linear operators can be found.

• See this for the existence of an unbounded linear functional on an infinite dimensional space. You can define an unbounded linear operator from this. Dec 15 '17 at 7:35

We have no examples of unbounded linear operators between these spaces. The Axiom of Choice tells us there are many such operators, but it does not tell us what they are, so we don't have any way to write down such an operator.

Wikipedia goes into this subject too:

As noted above, the axiom of choice (AC) is used in the general existence theorem of discontinuous linear maps. In fact, there are no constructive examples of discontinuous linear maps with complete domain (for example, Banach spaces). [...] it is consistent with set theory without AC that there are no discontinuous linear maps on complete spaces. In particular, no concrete construction such as the derivative can succeed in defining a discontinuous linear map everywhere on a complete space.

With AC, the proof of existence of an unbounded linear map $T:X\to Y$, for any $X, Y$, can proceed like this: let $\{v_\alpha\}$ be a Hamel basis of $X$ and pick any nonzero vector $w\in Y$. Take any unbounded collection of numbers $\{c_\alpha\}$ indexed by the same set as $\{v_\alpha\}$. Define $T(v_\alpha) = c_\alpha \|v_\alpha\| w$ and extend by linearity (which is possible since $\{v_\alpha\}$ is a basis).

The above does not produce an example of an unbounded linear operator on an infinite-dimensional Banach space because there are no examples of a Hamel basis for such a space.

• I would consider a normalized Hamel basis to keep the vector norms, in the basis, under control. Dec 4 '21 at 20:42

$1.$ Let $T : \ell^{\infty} \rightarrow \ell^{\infty}$ be given by $T((x_1, x_2, , x_3\ldots) = (1 \cdot x_1, 2 \cdot x_2, 3 \cdot x_3, \ldots)$.

Then, for $e_{n} = (0, \ldots,0, 1, 0,\ldots, 0, \ldots)$ $n \in \mathbb{N}$, (where $1$ appears at the $n$th position), we get $T(e_{n}) = n e_{n}$. Now $\|T(e_{n})\|_{\infty}= n \|e_{n}\|_{\infty} = n \cdot 1 = n.$ Now as $n \rightarrow \infty$, the sequence of real numbers $\|T(e_n)\|$ goes to $\infty$. So, $\|T\| = \sup \{ \|T(x)\|_{\infty}: \|x\|_{\infty} =1 \} = \infty.$

$2.$ Let $T : \ell^{p} \rightarrow \ell^{p}$ be given by $T((x_1, x_2, , x_3\ldots) = (1 \cdot x_1, 2 \cdot x_2, 3 \cdot x_3, \ldots)$.

Then, for $e_{n} = (0, \ldots,0, 1, 0,\ldots, 0, \ldots)$ $n \in \mathbb{N}$, (where $1$ appears at the $n$th position), we get $T(e_{n}) = n e_{n}$. Now $\|T(e_{n})\|_{p}= \| (0, \ldots, 0, n, 0, \ldots)\|_{p} = (n^p)^{1/p}=n$. Now as $n \rightarrow \infty$, the sequence $\|T(e_n)\|_{p}$ goes to $\infty$. So, $\|T\| = \sup \{ \|T(x)\|_{p}: \|x\|_{p} =1 \} = \infty.$

$3.$ Same example works for an unbounded operator between $\ell^p$ to $\ell^{\infty}$ and vice-versa.

• These operators are not defined on the whole spaces $l^p$ or $l^\infty$.
– daw
Dec 15 '17 at 12:43
• @ShubhamNamdeo thank you for your answer, but how to show that for every $x \in \ell^\infty$ or $x \in \ell^p$, the image under your $T$ of that $x$ is actually in $\ell^\infty$ or $\ell^p$, respectively? Dec 15 '17 at 13:48
• Thank you @SaaqibMahmuud for pointing out my mistake! Jan 1 '18 at 10:43