Gradient of inner product in Hilbert space

Let $\mathcal{H}$ be a Hilbert space and \begin{align} f&\colon \mathcal{H} \to \mathbb{R}\\ f(x) &= ||x-c||_\mathcal{H} ^2 \end{align} from some constant $c \in \mathbb{H}$

Is the derivative of $f$ at $x$ equal $2x-2c$?

I used \begin{align} f(x) &= \langle x,x \rangle - 2\langle x,c \rangle + \langle c,c \rangle \end{align} and the standard rules for derivation

\begin{align} \frac{\delta}{\delta x} f &= \frac{\delta}{\delta x} \langle x,x \rangle - \frac{\delta}{\delta x} 2\langle x,c \rangle + \frac{\delta}{\delta x} \langle c,c \rangle \\ &= 2x - 2 c \end{align}

Are there any special "obstacles" I have to consider or can I just use the these rules as in the standard euclidean space?

• if your Hilbert space is real, then everything is fine. – user251257 Jul 21 '15 at 7:39
• I would say that the derivative of $f$ is the linear map $Df(x) \colon h \mapsto 2\langle x-c\mid h \rangle$ and that $\nabla f(x) = 2(x-c)$. But you are right, up to Riesz representation theorem. – Siminore Jul 21 '15 at 8:15