I feel like this has to have been asked before, but my searches turned up nothing.

I was tutoring a student today and they asked me what is a good intuition for an adjoint. I still am not sure I know the answer to that question, but I figured I would think it through. I know there is a strong relationship between inner products and adjoints (by definition) and inner products and dual spaces (by Riesz).

However, it occurred to me as I was trying to sort this out in my mind: I have no idea what is a good definition of the inner product on a dual space. That is, if $V$ is a Hilbert space, how do we use the inner product of $V$ to create an inner product on $V^*$? Surely one must exist, but I couldn't think up an obvious example, except by passing through Riesz. Is there something more elementary?

[Side Note: I'm not sure how much I like adjoint as a tag, but okay]


You can define the inner product without passing through Riesz. You can define the norm of functionals using the definition of operator norm. Once you have the norm you can use the polarization identity to define the inner product.

The resulting formulas are:

If $x^*,y^*\in H^*$ then you can define $$(x^*,y^*)=\frac{1}{4}\left(\sup_{\|x\|=1,\ x\in H}|x^*(x)+y^*(x)|^2-\sup_{\|x\|=1,\ x\in H}|x^*(x)-y^*(x)|^2\right)$$

for the case of a real Hilbert space and

$$(x^*,y^*)=\frac{1}{4}\left(\sup_{\|x\|=1,\ x\in H}|x^*(x)+y^*(x)|^2-\sup_{\|x\|=1,\ x\in H}|x^*(x)-y^*(x)|^2+i\sup_{\|x\|=1,\ x\in H}|x^*(x)+iy^*(x)|^2-i\sup_{\|x\|=1,\ x\in H}|x^*(x)-iy^*(x)|^2\right)$$

for the case of a complex inner product.

But in the question you talk about many things, so I am not sure which you needed more. Insight about the adjoint the above is not.

The two ideas that I find more useful for this are:

  1. The definition, together with the notation $(x,x^*)$ to denote $x^*(x)$. This notation makes the dual elements acting on elements of the space look like inner product and adjoints behave by jumping form one side to another of the pair symbol.

  2. The geometric interpretation for the finite dimensional case.

  • $\begingroup$ Yeah, I see how that goes. Thank you :) $\endgroup$ – Eric Stucky Mar 1 '15 at 19:29

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