This is a seeming "paradox" that has been bothering me for some time, as it (or other situations like it) show up often when computing gradients for numerical optimization on complex vector spaces. Today I finally sat down to try to clear the confusion, but after much time being confused and going in circles, I still haven't been able get a good intuition for what's going on.
The basic issue is understanding how exactly the derivative of a function $f$ doesn't have a Riesz representative when viewed on a complex vector space $\mathbb{C}^n$, whereas the exact same function's derivative does have a Riesz representative when viewed on the real vector space $\mathbb{R}^{2n}$.
To that end, let $f:\mathbb{C}^n \rightarrow \mathbb{C}$, $$f(u) = u^*u$$ and consider the Gauteaux derivative of $f$ at $u$ in the direction $h$, \begin{align} f'(u)h &= \lim_{s \rightarrow 0}\frac{1}{s}[f(u+sh) - f(u)] \\ &= u^*h + h^*u. \end{align}
The natural question is: does $f'(u)$ have a Riesz representative (gradient vector) $g$ such that $(g,h) = f'(u)h$ for all $h$?
In the complex inner product $(v,w):=v^*w$, the answer must be no, since $u^*h + h^*u$ is always real, whereas $(g,ig)$ would be complex for any candidate gradient vector $g$.
On the other hand, breaking $u = u_R + i u_I$ and $h = h_R + i h_I$ into their constituent real and imaginary parts yields, \begin{align} u^*h + h^*u &= (u_R - i u_I)^T(h_R + i h_I) + (h_R - i h_I)^T(u_R + i u_I) \\ &= u_R^T h_R + u_I^Th_I \\ &= \begin{bmatrix}u_R^T & u_I^T\end{bmatrix}\begin{bmatrix}h_R \\ h_I\end{bmatrix} \end{align}
And so $f'(u)$ does have a Riesz representative, $$g = \begin{bmatrix}u_R \\ u_I\end{bmatrix}$$ in the inner product space $\mathbb{R}^{2n}$.
So what's going on here? The calcuations seem bulletproof, but the conclusion is unsettling. What is the intuition for why $f'(u)$ has Riesz representative on the real vector space $\mathbb{R}^{2n}$, but no Riesz representative on $\mathbb{C}^n$?
There is no contradiction because your $f'(u)$ is not a linear function over $\mathbb{C}$.
EDIT. In fact, $\lim_{s \rightarrow 0}\frac{1}{s}[f(u+sh) - f(u)]$ does not exist. Indeed, this limit (where $s$ is a complex number !) exists only if $f$ is a holomorphic function; that is not the case here, in particular, because $\lim_{s\rightarrow 0}\dfrac{\bar{s}}{s}$ does not exist.