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Let $X,Y$ be normed vector spaces, X infinite dimensional. Can I find 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?

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up vote 21 down vote accepted

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 $D$.

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Yeah, nice! And it answers my questions: $\ker f$ is always a subspace. – Matthew Daws Jun 16 '10 at 21:15
It looks like the best possible answer! – Pierre-Yves Gaillard Jun 17 '10 at 3:45
Wonderful! I choose this as answer for simplicity and for the fact that it works in every infinite dimensional normed space – Nicolò Jun 17 '10 at 10:18
I wrote a tiny complement to this answer:… – Pierre-Yves Gaillard Jun 18 '10 at 4:50

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.

I think I have NOT used the Axiom of Choice.

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Nice example. One could also use Stone-Weierstrass in place of Müntz-Szász: the even polynomials are an algebra that separates points and contains the constants. – Nate Eldredge Jun 16 '10 at 19:43

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 $X$.

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How do you guarantee the existence of such $Z$? If, say, $X$ is a separable Banach space, you could take $Z$ to be the algebraic span of a countable dense subset, and by the Baire category theorem it has uncountable algebraic codimension. But I don't see what to do if $X$ is incomplete (and possibly of countable algebraic dimension), or if $X$ is complete but not separable. – Nate Eldredge Jun 16 '10 at 18:18
Nate, there are certainly examples of $X$ which work, as you say, and these include many of the standard examples. For general normed spaces, I'm not sure. I would suspect that the intersection of the kernels of a sequence of linearly independent unbounded functionals might work, but whether there do exists such sequences, I'm not sure... – Robin Chapman Jun 16 '10 at 18:26
You don't need infinite codimension. Any nontrivial dense subspace does the job (codimension 1 is enough). And it is easy to construct a dense subsbace: find one in some closed subspace (e.g. a separable one), then add a complementary subspace. – Sergei Ivanov Jun 16 '10 at 19:50

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!

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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.

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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.

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Your argument shows that $T |_D$ has a bounded extension to all of $X$, but this bounded extension need not be the same as $T$. – Mark Meckes Jun 16 '10 at 19:11
Not quite. The extension is continuous, but what if that extension is not the original map? Can't you have two linear maps that agree on a dense set, but differ off that dense set? – Gerald Edgar Jun 16 '10 at 19:15
For a concrete example: let X be the subspace of $c_0$ spanned by, say, the sequence $x_0=(1/n)$ and all the sequences which are eventually zero (that latter being the subspace $c_{00}$). Then you can define a linear map $T:X\rightarrow\mathbb R$ to be zero on $c_{00}$ and anything you like on $x_0$. But $c_{00}$ is dense in $X$. Your continuous extension is just the zero map! – Matthew Daws Jun 16 '10 at 19:25
Thanks guys, I understand the flaw in my reasoning now. – Tom LaGatta Jun 16 '10 at 20:47

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