Let $F[h]$ be a generic functional of the function $h(x):{\mathbb{R}^d}\mapsto \mathbb{R}$. Consider the following definition of Gateaux derivative in direction $g(x):{\mathbb{R}^d}\mapsto \mathbb{R}$
\begin{equation} \tag{1} \frac{\partial F[h]}{\partial g}\triangleq \left[\frac{\text{d}}{\text{d}\varepsilon}F[h+\varepsilon g]\right]_{\varepsilon=0} \end{equation}
and the following definition for the second order derivative (still in direction $g(x)$)
\begin{equation} \tag{2} \frac{\partial^2 F[h]}{\partial g^2}\triangleq \left[\frac{\text{d}^2}{\text{d}\varepsilon^2}F[h+\varepsilon g]\right]_{\varepsilon=0} \end{equation}
given these defintions, I would like to prove (or disprove) the following relation \begin{equation} \frac{\partial^2 F[h]}{\partial g^2}=\frac{\partial}{\partial g}\left[\frac{\partial F[h]}{\partial g}\right] \tag{$\star$} \end{equation}
My attempt
Before starting, note that from $(1)$ follows immediately the simplified formula for the first order Gateaux derivative \begin{equation} \tag{3} \frac{\partial F[h]}{\partial g}=\lim_{\tau \to 0} \frac{F[h+\tau g]-F[h]}{\tau} \end{equation} Now, my strategy is to show that the RHS of $(\star)$ and the LHS of $(\star)$ are equal to a common expression. So, for the RHS we have from $(3)$ that \begin{equation*}\begin{aligned} \frac{\partial}{\partial g}\left[\frac{\partial F[h]}{\partial g}\right]&= \frac{\partial}{\partial g}\left[\lim_{\tau \to 0} \frac{F[h+\tau g]-F[h]}{\tau}\right]\\ &=\lim_{\tau \to 0} \frac{1}{\tau} \left\{ \frac{\partial F}{\partial g}[h+\tau g]-\frac{\partial F}{\partial g}[h]\right\}\\ &=\lim_{\tau \to 0} \frac{1}{\tau} \left\{ \lim_{\sigma\to 0}\frac{F[(h+\sigma g)+\tau g]-F[h+\tau g]}{\sigma}-\frac{F[h+\sigma g]-F[h]}{\sigma}\right\}\\ &=\lim_{\tau,\sigma \to 0} \frac{F[h+(\sigma+\tau) g]-F[h+\tau g]-F[h+\sigma g]+F[h]}{\tau\,\sigma} \end{aligned} \end{equation*}
For the LHS we have to compute $(2)$. In order to do it, note that \begin{equation*}\begin{aligned} \frac{\text{d}^2}{\text{d}\varepsilon^2}F[h+\varepsilon g]&=\frac{\text{d}}{\text{d}\varepsilon}\left[\frac{\text{d}}{\text{d}\varepsilon}F[h+\varepsilon g]\right]\\ &=\frac{\text{d}}{\text{d}\varepsilon}\left[\lim_{\tau \to 0} \frac{F[h+(\varepsilon+\tau)g]-F[h+\varepsilon g]}{\tau}\right]\\ &=\frac{\text{d}}{\text{d}\varepsilon}\left[\lim_{\tau \to 0} \frac{F[h+\varepsilon g+\tau g]-F[h+\varepsilon g]}{\tau}\right]\\ &=\lim_{\tau \to 0} \frac{1}{\tau}\left[\frac{\text{d}F}{\text{d}\varepsilon}[h+\varepsilon g+\tau g]-\frac{\text{d}F}{\text{d}\varepsilon}[h+\varepsilon g]\right]\\ &=\lim_{\tau \to 0} \frac{1}{\tau}\left[\lim_{\sigma \to 0} \frac{F[h+(\varepsilon+\sigma) g+\tau g]-F[h+\varepsilon g+\tau g]}{\sigma}-\frac{F[h+(\varepsilon+\sigma) g]-F[h+\varepsilon g]}{\sigma}\right]\\ &=\lim_{\tau,\sigma \to 0} \frac{F[h+(\varepsilon+\sigma) g+\tau g]-F[h+\varepsilon g+\tau g]-F[h+(\varepsilon+\sigma) g]+F[h+\varepsilon g]}{\tau \sigma} \end{aligned}\end{equation*} Accordingly, the LHS is given by \begin{equation*}\begin{aligned} \frac{\partial^2 F[h]}{\partial g^2}&=\left[\frac{\text{d}^2}{\text{d}\varepsilon^2}F[h+\varepsilon g]\right]_{\varepsilon=0}\\ &=\lim_{\tau,\sigma \to 0} \frac{F[h+\sigma g+\tau g]-F[h+\tau g]-F[h+\sigma g]+F[h]}{\tau \sigma} \end{aligned}\end{equation*} which is the same as the RHS.
Question
I'm not sure if my proof is correct because I don't know if I have written correctly the limits involved in the definitions of the derivatives. So my question is the following: do you see any problem in my derivation of $(\star)$?