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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?

Community
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Mathems
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2 Answers2

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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.

hmakholm left over Monica
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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.

copper.hat
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