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 $$ \lVert T \rVert = \infty. $$

  • $\begingroup$ @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. $\endgroup$ – Saaqib Mahmood Apr 13 '20 at 19:02
  • $\begingroup$ @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. $\endgroup$ – Saaqib Mahmood Apr 13 '20 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 $X\to X$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.