# Does boundedness of all linear operators implies domain space is finite dimensional?

Theorem: If a normed space $$X$$ is finite dimensional then every linear operator on $$X$$ is bounded.

I have a proof to this. I was thinking about the converse "If every linear operator on normed space $$X$$ is bounded then $$X$$ is finite dimensional."

My question is: "Is the converse true?" My guess is NO. But I am not getting a counter example.

What you are looking for is whether or not the following holds:

Let $$X$$ and $$Y$$ be normed spaces, either both real or both complex, such that $$\dim X = \infty$$ and such that $$Y$$ has non-zero vectors. Then there exists an unbounded linear operator $$T \colon X \rightarrow Y$$.

Let us attempt a proof.

Let $$X$$ be an infinite-dimensional real normed space.

Let $$S$$ be a linearly independent, countably infinite, ordered subset of $$X$$, say, $$S = \left\{ \, x_1, x_2, x_3, \ldots \, \right\}.$$ This $$S$$ can be extended to a basis (in fact an ordered basis) $$B$$ for $$X$$, using the choice axiom.

Now let us define a linear operator $$T \colon X \rightarrow \mathbb{R}$$ as follows: $$T\left( x_n \right) \colon= n \qquad T (x) \colon= 0 \ \mbox{ if } x \in B \setminus S.$$ Of course, a linear operator is uniquely determined by its values at the elements of a basis of its domain.

Now let us assume without any loss of generality that $$\lVert x_n \rVert_X = 1$$ for every $$n \in \mathbb{N}$$.

Then for any $$n \in \mathbb{N}$$, we have $$\lVert T \rVert \geq n.$$ Therefore $$\lVert T \rVert = \infty.$$

Now let us generalise this.

Let $$X$$ and $$Y$$ be normed spaces, either both real or both complex, such that $$\dim X = \infty$$ and such that $$Y$$ has elements other than the zero vector $$\mathbf{0}_Y$$.

Let $$B$$ be an ordered basis for $$X$$ such that $$B$$ contains a countably infinite subset $$S = \left\{ \, x_1, x_2, x_3, \ldots \right\}$$ such that $$\left\lVert x_n \right\rVert_X = 1$$ for all $$n \in \mathbb{N}$$. And, let $$y_o$$ be a non-zero vector in $$Y$$.

Let us now define $$T \colon X \rightarrow Y$$ as follows: $$T \left( x_n \right) \colon= n y_0 \qquad T(x) = \mathbf{0} \ \mbox{ if } x \in B \setminus S.$$ Then $$\lVert T \rVert = \infty.$$

• @Ravi what I've shown above is as follows: Given an infinite-dimensional normed space, we can always define an unbounded linear operator with that normed space as the domain. Therefore we can conclude that if every linear operator on a particular normed spces is bounded, then that normed space is necessarily finite-dimensional. This is the converse you wanted to have, isn't it? Hope this now clarifies your confusion. Apr 13, 2020 at 19:02
• @Ravi no problem. You're welcome to contacting me through WhatsApp should you have any queries that I can answer orally. My contact info is in my profile. Apr 13, 2020 at 19:08

The converse is true: If $$X$$ is infinite-dimensional, then one can construct an unbounded linear functional on it. By identifying the scalars with a one-dimensional subspace of $$X$$, we obtain an unbounded linear map $$X\to X$$.