I have a function $f$ that is Hardamard differentiable but it is unclear to me if that also implies that $f$ is strictly differentiable according to Clarke (see Optimization and Nonsmooth Analysis).
Hardamard Derivative: A function $f:\mathbb{R}^N\to\mathbb{R}$ is called Hadamard directionally differentiable at a point $x\in\mathbb{R}^N$ if for any $g\in\mathbb{R}^N$ there exists the finite limit $$f'_{H}(x,g)=\lim_{[\alpha,g'\to [+0,g]]}\frac{f(x+\alpha g')-f(x)}{\alpha}.$$
Strict Derivative: We shall say that $F$ admits a strict derivative at $x$, an element of $\mathfrak{L}(X,Y)$ denoted $D_{s}F(x)$, provided that for each $v$. the following holds: $$\lim_{x'\to x, \,\,t\downarrow 0}\frac{F(x'+tv)-F(x')}{t}=< D_{s}F(x),v >.$$
Note: we are working in $\mathbb{R}^N$, where $F$ is assumed to be Lipschitz.
These two definitions seem very similar to me but I am unsure how to relate them. Any insight and/or guidance on this will be greatly appreciated (:
So, what I have found so far is that if the Banach space is finite $X=\mathbb{R}^n$, then the Hadamard derivative and the Frechet derivative coincide. Also, if $f$ is Lipschitz, then the Hadamard derivative coincides with the Gateaux derivative.
Next, if $f$ is continuously Frechet differentiable at $x$ then $f$ is strictly differentiable.
So with that, it seems that if we are working in $\mathbb{R}^n$ and $f$ is continuously Hadamard differentiable at $x$, then $f$ is strictly differentiable at $x$.