Unbounded operator bounded in a dense subset

Question

Let $X, Y$ be normed vector spaces, where $X$ is infinite dimensional. Does there exist a linear map $T : X \rightarrow Y$ and a subset $D$ of $X$ such that $D$ is dense in $X$, $T$ is bounded in $D$ (i.e. $\sup _{x\in D, x \neq 0} \frac{\|Tx\|}{\|x\|}<\infty $), but $T$ is not bounded in $X$?

Answer 1

Matthew's answer reminded me of a fact that makes this easy: if $X$ is a normed space (say, over $\mathbb{R}$) and $f : X \to \mathbb{R}$ is a linear functional, then its kernel $\ker f$ is either closed or dense in $X$, depending on whether or not $f$ is continuous (i.e. bounded). The proof is trivial: $\ker f$ is a subspace of $X$ of codimension 1. Its closure is a subspace that contains it, so must either be $\ker f$ or $X$. And of course, a linear functional is continuous iff its kernel is closed. This is Proposition III.5.2-3 in Conway's A Course in Functional Analysis.

So let $f$ be an unbounded linear functional on $X$ (which one can always construct as in Matthew's example), and take $D = \ker f$. $D$ is dense by the above fact, and $f$ is identically zero on $D$.

Answer 2

Let $X$ be the space of real polynomials, normed as functions in $C[a,b]$. Here we want $0 < a < b$ fixed. Now define $T \colon X \to X$ so that $T(x^n) = 0$ if $n$ is even and $T(x^n) = nx^n$ if $n$ is odd. Then $T$ is unbounded, but it vanishes on the set of even polynomials. That set is dense by the Müntz-Szász theorem. http://en.wikipedia.org/wiki/M%C3%BCntz%E2%80%93Sz%C3%A1sz_theorem

I think I have NOT used the Axiom of Choice.

Answer 3

Take a dense linear subspace $Z$ of $X$ such that $X/Z$ is infinite-dimension (algebraically). Let $x_1,x_2,\ldots$ be a sequence of elements of $X$ whose images in $X/Z$ are linearly indepdendent. We can define a linear map on the algebraic span on $Z$ and the $x_j$ which is zero on $Z$ and satisfies $\|Tx_n\|=n\|x_n\|$. Extend this to all $X$ by Zorn's lemma. Then $T$ is zero on the dense space $Z$ but unbounded on $X$.

Answer 4

X is infinite-dimensional, so we can find $(e_n)$ a linearly independent sequence in X; let X' be the span. By rescaling, we can assume that $\|e_n\|=1$ for each n. Define $T:X'\rightarrow \mathbb R$ (or $\mathbb C$, or embed into Y if you wish) by $T(e_n) = n$ for each n. Clearly T is unbounded on X'.

For each finite sum $x=\sum_{n=1}^N x_n e_n$ and $\epsilon>0$, we can choose $a\in\mathbb R$ and $m$ very large with $|a|<\epsilon$ and $T(x) = -am$. Set $y=x+ae_m$, so $T(y)=T(x)+am=0$ and also $\|x-y\| = |a|<\epsilon$. Let D' be the collection of all such $y$; as such $x$ exhaust X', we certainly have that D' is dense in X'.

Use Zorn to extend $E=\{e_n\}$ to $E'$ a basis of X. Extend T to X by setting $T(x)=0$ for $x\in E' \setminus E$. Let $D = D' + \text{span}(E'\setminus E)$, so D is dense in X, and T is bounded on D; actually T vanishes on D.

Now, D is certainly not a subspace: if you want that as well, I don't know!

Answer 5

This is just to make Nate Eldridge's answer selfcontained.

For any normed vector space $V$ and any $r > 0$, write $V_r$ for the open ball of radius $r$ and center $0$ in $V$.

Let $X$ be a normed vector space, $Y$ a closed space, $Z$ the quotient, $\pi$ the canonical projection, $\tau$ the quotient topology, $\nu$ the topology on $Z$ induced by the quotient norm $$|\pi(x)|:=\inf_{y\in Y}|x+y|.$$

(It's easy to see that this is a norm.)

We claim $\tau=\nu$.

Both topologies are translation invariant.

The set $\{\pi(X_r)\ |\ r > 0\}$ is a basis for the $\tau$-neighborhoods of $0$ in $Z$.

The set $\{(Z_r)\ |\ r > 0\}$ is a basis for the $\nu$-neighborhoods of $0$ in $Z$.

As $\pi(X_r)=Z_r$, we're done.

Answer 6

No. Bounded means continuous, for linear operators. If a linear operator is bounded (continuous) on a dense subset, then you can extend it continuously to the whole space, which means it is bounded.