I have stumbled upon this interesting exercise, and I cannot seem to find conditions that satisfy all the properties needed, I can find several combinations which satisfy one or two of the properties, but not one for all three, I am really interested to see the solution so any help or hints would be greatly appreciated!
So the question is as follows: Give an example of vector spaces $\mathbb{X}$ and $\mathbb{Y}$, two norms $||\cdot||_a$ and $||\cdot||_A$ on $\mathbb{X}$, where the two norms are equivalent, two norms $||\cdot||_b$ and $||\cdot||_B$ on $\mathbb{Y}$, where the two norms are equivalent, and a function $f:\mathbb{X} \to \mathbb{Y}$ such that all of the following hold:
a) $\:f:(\mathbb{X},||\cdot||_a ) \to (\mathbb{Y},||\cdot||_b)$ is Frechet differentiable at every $x\in\mathbb{X}$;
b) $\:f:(\mathbb{X},||\cdot||_A ) \to (\mathbb{Y},||\cdot||_b)$ is not Frechet differentiable at any $x\in\mathbb{X}$;
c) $\:f:(\mathbb{X},||\cdot||_a ) \to (\mathbb{Y},||\cdot||_B)$ is not Frechet differentiable at any $x\in\mathbb{X}$.
It should be noted that the first part of this exercise was to prove that if a function is freshet differentiable everywhere with respect to two certain norms then if these are changed for equivalent ones the function is still freshet differentiable with the same derivative. So this can be assumed in the answer.
Thanks!
Theorem. Let $\mathrm{V}$ and $\mathrm{W}$ be two normed spaces, and let $\| \cdot \|_1$ and $\| \cdot \|_a$ be a two equialent norms in $\mathrm{V}$ and, likewise $\| \cdot \|_2$ and $\| \cdot \|_b$ is a pair of equivalent norms for $\mathrm{W}.$ Suppose $f$ is defined on an open subset $\mathrm{A}$ of $\mathrm{V}$ into $\mathrm{W}$ and let $v$ be a point of $\mathrm{A}.$ Then, $f$ is differentiable in $v$ with respect to the pair $(\| \cdot \|_1, \| \cdot \|_2)$ if and only if it is differentiable in $v$ with respect to the pair $(\| \cdot \|_a, \| \cdot \|_b).$
Proof. There are constants $c, d, p, q$ all positive such that $$c \| \cdot \|_1 \leq \| \cdot \|_a \leq d \| \cdot \|_1$$ and $$p \| \cdot \|_2 \leq \| \cdot \|_b \leq q \| \cdot \|_2.$$ Then $$\dfrac{c}{q} \dfrac{\|f(v + h) - f(v) - Th\|_b}{\|h\|_a} \leq \dfrac{\|f(v + h) - f(v) - Th\|_2}{\|h\|_1} \leq \dfrac{d}{p} \dfrac{\|f(v + h) - f(v) - Th\|_b}{\|h\|_a}$$ and the proof is complete.
How this theorem imply the excercise? Easy, clearly $\| \cdot \|$ is equivalent with itself.