Reference: A primer of nonlinear analysis - Antonio & Giovanni
Let $H$ be a hilbert space over $\mathbb{K}$ and $U$ be open in $H$ and $p\in U$ and $f:U\rightarrow \mathbb{K}$ be a functional Fréchet differentiable at $p$.
By Riesz Representation theorem, there exists a unique $y\in H$ such that $\forall x\in H$, $Df(p)(x)=<x,y>$.
Is $y$ the gradient of $f$ at $p$?
No, it is not. The reason why is that, if $\mathbb{K}$ is complex, then the inner product is conjugate linear and therefore only induces an anti-isomorphism, not an isomorphism, between $H$ and its dual. We find that $y$ is identified with the conjugate of the gradient in this case.
To be more concrete, pick $H = \mathbb{C}$ (over $\mathbb{C}$). Then pick the linear function $f(z) = iz$. Then, at any point, the derivative map of $f$ is itself. But, the unique complex number $y$ such that $\langle z, y \rangle = iz$ is given by $y = -i$, if we choose the convention that we conjugate in the second argument. So then $y$ would be identified with the linear map that multiplies by $-i$, which is the conjugate of what we normally call the gradient.