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?


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.


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.


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.