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.
 A: 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 $X\to X$.
A: 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 
  $$ \lVert T \rVert = \infty. $$

