Let $X$ and $Y$ be normed spaces, either both real or both complex; let $T \colon X \to Y$ be a linear mapping. If there is a real number $r > 0$ such that $$ \lVert T(x) \rVert_Y \leq r \lVert x \rVert_X \qquad \mbox{ for all } x \in X, $$ then $T$ is said to be a bounded linear operator.

One example of this operator is the following:

Let $X$ denote the normed space of all the real polynomials defined everywhere on the closed interval $[0, 1]$, with the maximum norm, and let $T \colon X \to X$ be the map $x \mapsto x^\prime$. Then this linear operator is unbounded.

Can we find any other examples of unbounded linear operators?

I know that every linear operator whose domain is a finite-dimensional normed space is bounded.

  • $\begingroup$ In $\ell^2$, send the $n$th standard unit vector $\mathbf{e}_n$ to $n\mathbf{e}_n$ and extend linearly. $\endgroup$ – max_zorn Dec 28 '17 at 6:52
  • $\begingroup$ @max_zorn, this is the same as differentiation via Fourier series. $\endgroup$ – fourierwho Dec 28 '17 at 8:05
  • $\begingroup$ How so, fourierwho? $\endgroup$ – max_zorn Dec 28 '17 at 8:34
  • $\begingroup$ Not quite the same. Taking Fourier series in reverse has the effect of sending $e_n \in \ell^2(\mathbb{Z})$ to $\exp(2 \pi i n x) \in \ell^2(S^1)$, and taking the derivative multiplies this by $2 \pi i n$. $\endgroup$ – Qiaochu Yuan Dec 28 '17 at 10:34
  • $\begingroup$ In any case, writing down unbounded operators is easy. Consider any operator $T$ defined on polynomials by sending $x^n$ to $r_n x^n$. This operator is bounded iff the sequence $r_n$ is bounded (exercise). $\endgroup$ – Qiaochu Yuan Dec 28 '17 at 10:35

First a word of caution: If people speak about an unbounded operator from $X$ to $Y$, they mostly mean a (unbounded) linear map $T\colon D \rightarrow Y$, where $D$ is linear subspace of $X$ (see here). This allows to deal with differential operators in a context which is more general than that of polynomials. For example take $X=Y=L^2[0,1]$, $D=C^1[0,1]$ and $Tx=x'$ as before.

Of course you can always extend such a map $T\colon D \rightarrow Y$ to a linear map $\tilde T\colon X \rightarrow Y$ by simply setting $\tilde T\vert_{\tilde D} = 0 $ on some complementary subspace $\tilde D$ (i.e. a subspace with $X= D \oplus \tilde D$). This will also be unbounded, but does neither contain any new information nor respect the norm of $X$.

Using this notion of unbounded operators, at least on function spaces there is a huge class of examples, which generalizes differential operators. For the sake of simplicity let us consider $X=Y=L^2\mathbb{R}$.

1) Multiplication operators: Given $m\in L^1_{\text{loc}}\mathbb{R}$, put $D=\{x\in L^2\mathbb{R}\vert ~mx\in L^2\mathbb{R}\}$ and $Tx= mx$ for $x\in D$. This is a bounded operator iff $m$ is bounded.

2) Fourier Multiplication operators: The Fourier transform is a bounded linear map $F\colon L^2\mathbb{R} \rightarrow L^2\mathbb{R}$ (even an isometric isomorphism if you use the right normalization) and can be used to generate operators of the form $$ Tx=F^{-1}\left(m\cdot(Fx)\right), \quad D=\{x\in L^2\mathbb{R}\vert~F^{-1}\left(m\cdot(Fx)\right) \in L^2\mathbb{R}\} $$ Since $\Vert Tx \Vert = \Vert m \cdot (Fx)\Vert $ and $\Vert Fx\Vert = \Vert x \Vert$, this will again be bounded iff $m$ is a bounded function. Taking $m(\xi)=i\xi$ gives as assoicated operator exactly the differentiation operator $Tx=x'$.

3) Pseudodifferential operators: Take the same definition of $T$ as before, but allow $m$ to both depend on the spatial variable $t$ and the Fourier variable $\xi$. This generalizes the concepts above, $m(t,\xi)=m(t)$ gives a multiplication operator and $m(t,\xi)=m(\xi)$ gives a Fourier multiplication operator (see here for details).


If $X$ is an infinite-dimensional normed space, then you can choose a Hamel basis $\{ x_{\alpha} \}$ and define a linear functional $F$ by setting $F(x_{\alpha})=s_{\alpha}$, where $\{ s_{\alpha} \}$ is a set of scalars. $F$ will be unbounded if $\sup_{\alpha} |s_{\alpha}|/\|x_{\alpha}\|=\infty$. Then you can extend the argument to obtain $T: X \rightarrow Y$ by choosing a finite set of discontinuous functionals $F_n$ on $X$, choosing an independent set of vectors $\{ y_n \} \subset Y$, and then setting $F(x)=\sum_n F_n(x)y_n$.

  • $\begingroup$ thank you for your answer, but I would be even more grateful if you could please expand your post by elaborating upon the claims made therein. For example, what a Hamel basis is, how to show that our $F$ is indeed unbounded? $\endgroup$ – Saaqib Mahmood Jan 1 '18 at 10:59
  • $\begingroup$ A Hamel basis is a maximal linearly independent set. Every $x\in X$ can be written as a finite linear combination of elements in the Hamel basis. To show that $F$ is unbounded, use Hahn-Banach to find a continuous linear functional $\phi_n$ such that $\phi_n(y_k)=\delta_{n,k}$ and conclude that $\phi\circ F = F_n$ is unbounded, which would force $F$ to be unbounded. $\endgroup$ – DisintegratingByParts Jan 1 '18 at 15:22

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