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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.

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

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  • $\begingroup$ @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
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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 $X\to X$.

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