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I'm reading a paper and I wonder about the statement:

$f'(x)h=(y(x),h)_{L^2}$ then follows with the Riesz isomorphism that $f'(x)=y(x)$

$f'(x)h$ is the Gateaux derivative. I don't see that with the Riesz representation theorem. Can someone explain that?

Thanks!

brian93
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1 Answers1

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You do not provide much context, therefore I can only guess that $f'(x)$ is a linear and continuous functional, so by the Riesz representation theorem there must exist a vector $y(x) \in L^2$ such that $f'(x) = \langle y(x), \cdot \rangle$.

Note, though, that the Gâteaux derivative need not be linear or continuous. Nevertheless, if $f$ is Gâteaux-derivable over the $complex$ numbers, then it is Fréchet derivable and everything turns out nicely then. (Therefore, I assume your $L^2$ to be a complex space.)

Alex M.
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