How do i prove that every linear operator between finite-dimensional Hilbert spaces is bounded? When I learned basic linear-algebra, "adjoint" was only defined for linear operator between finite-dimensional inner product spaces.
Right now, I'm studying Hilbert spaces and I want the past definition consistent with a new definition.
I have proved following theorem in basic linear-algebra:

Let $V,W$ be inner product spaces over $\mathbb{F}$.
Let $T:V\rightarrow W$ be a linear operator.
If $V$ is finite-dimensional, there exists a unique function $T^*$ such that $\langle T(x),y\rangle=\langle x,T^*(y)\rangle$.

So my question is;
How do I prove that $T$ is bounded when $V$ and $W$ are finite-dimensional?
Moreover, is it true when $V$ is finite-dimensional but $W$ is not?
 A: Even if $W$ is not finite-dimensional, the range of $T$ will be a finite-dimensional subspace of $W$ (since it is spanned by the image of a basis for $V$), and that subspace is all that matters for the boundedness of $T$.
So without loss of generality we can assume $W$ is finite dimensional.
Choose orthonormal bases for $V$ and $W$, and write down the matrix for $T$. Then $\|T\|$ cannot possibly exceed the sum of the absolute values of the matrix entries.
A: Let $v_k$ be a basis for $V$. Since you are in a Hilbert space, you can assume that the basis is orthonormal.
If $v \in V$, let $x_k = \langle v_k, v \rangle$, then we have $\sum_k |x_k|^2 = \|v\|^2$, and letting $\|x\|_2^2 = \sum_k |x_k|^2$, we have $\|x\|_1 \le \sqrt{n} \|x\|_2$, where $n$ is the dimension of $V$.
Then $\|T v\| = \|T ( \sum_k x_k v_k ) \| \le \sum_k \|T v_k\| |x_k| \le K \|x\|_1 \le \sqrt{n}K \|x\|_2= \sqrt{n}K \|v\|$, where $K = \max_k \|T v_k\|$.
The same analysis applies in any finite dimensional normed space, except that one uses the fact that any two norms on such a space are equivalent, rather than the explicit relationship between $\|\cdot\|_1$ and $\|\cdot\|_2$ used above.
