Can we show $\lim_{\varepsilon\to0}\sup_{0<\left\|x-y\right\|_E<\varepsilon}\frac{|f(x)-f(y)|}{\left\|x-y\right\|_E}=\left\|{\rm D}f(x)\right\|_{E'}$?

Question

Let $f:E\to\mathbb R$ be Fréchet differentiable and $x\in E$. By definition, $$\lim_{y\to x}\frac{|f(x)-f(y)-{\rm D}f(x)(x-y)|}{\left\|x-y\right\|_E}=0.\tag1$$ Can we somehow (maybe by imposing a further suitable assumption) show that $$\lim_{\varepsilon\to0}\sup_{\substack{y\in E\\0<\left\|x-y\right\|_E<\varepsilon}}\frac{|f(x)-f(y)|}{\left\|x-y\right\|_E}=\left\|{\rm D}f(x)\right\|_{E'}\tag2?$$

Answer 1

If $E$ is a Banach space, and $u \colon E \to \mathbb{R}$ is a bounded linear functional, then for all $\delta > 0,$ $$ \|u\| = \sup_{\|x\| = 1}|u(x)| = \sup_{\|h\| = \delta} \left\lvert{u}\left(\frac{h}{\|h\|}\right)\right\rvert = \sup_{\|h\| = \delta}\frac{|u(h)|}{\|h\|}. $$ Therefore for all $\varepsilon > 0,$ \begin{equation} \label{3708359:eq:3}\tag{3} \|u\| = \sup_{0 < \|h\| < \varepsilon}\frac{|u(h)|}{\|h\|}. \end{equation}

If $U$ is an open subset of $E,$ and $f \colon U \to \mathbb{R}$ is differentiable at $x \in U,$ then the function $$ d \colon U - x \to \mathbb{R}, \ h \mapsto f(x + h) - f(x) - f'(x)(h) $$ satisfies the condition $$ \lim_{h \to 0}\frac{d(h)}{\|h\|} = 0. $$ Take any $\eta > 0,$ and take $\varepsilon > 0$ such that if $\|h\| < \varepsilon$ then $x + h \in U$ and $$ |d(h)| \leqslant \eta\|h\|. $$

If $x + h \in U,$ then by the Triangle Inequality, $$ ||f(x + h) - f(x)| - |f'(x)(h)|| \leqslant |f(x + h) - f(x) - f'(x)(h)| = |d(h)|. $$ Therefore, if $0 < \|h\| < \varepsilon,$ $$ \left\lvert \frac{|f(x + h) - f(x)|}{\|h\|} - \frac{|f'(x)(h)|}{\|h\|} \right\rvert \leqslant \eta. $$

Take any $\varepsilon^*$ such that $0 < \varepsilon^* < \varepsilon,$ and let $$ N(\varepsilon^*) = \{h \in E : 0 < \|h\| < \varepsilon^* \}. $$ For $h \in N(\varepsilon^*),$ the inequality just proved has the form $$ |p(h) - q(h)| \leqslant \eta, $$ for certain functions $$ p, q \colon N(\varepsilon^*) \to \mathbb{R}_{\geqslant 0}. $$ By \eqref{3708359:eq:3}, the function $q(h) = |f'(x)(h)|/\|h\|$ is bounded on $N(\varepsilon^*),$ and $$ \sup_{h \in N(\varepsilon^*)}q(h) = \|f'(x)\|. $$ Therefore, the function $p(h) = |f(x + h) - f(x)|/\|h\|$ is also bounded on $N(\varepsilon^*),$ and $$ \left\lvert \sup_{0 < \|h\| < \varepsilon^*} \frac{|f(x + h) - f(x)|}{\|h\|} - \|f'(x)\| \right\rvert \leqslant \eta. $$ This holds for arbitrary $\eta$ and for all $\varepsilon^*$ such that $0 < \varepsilon^* < \varepsilon,$ where $\varepsilon$ depends on $\eta,$ therefore $$ \lim_{\varepsilon \to 0+} \sup_{0 < \|h\| < \varepsilon} \frac{|f(x + h) - f(x)|}{\|h\|} = \|f'(x)\|. $$

Answer 2

Let $\{v_n\}$ be such that $\lim_{k\rightarrow \infty} \|Df(x)(v_n)\|=\|Df(x)\|$ and $\|v_n\|=1$ for all $n$. Then, by definition, choosing $t_n \rightarrow 0$, we have $$\lim_{n \rightarrow \infty}\frac{|f(x+t_n v_n)-f(x)|}{|t_n|}=\lim_{n\rightarrow \infty}|Df(x)(v_n)|=\|Df(x)\|.$$ Hence, $$\lim_{\varepsilon\to0}\sup_{\substack{y\in E\\0<\left\|x-y\right\|_E<\varepsilon}}\frac{|f(x)-f(y)|}{\left\|x-y\right\|_E} \geq \left\|{\rm D}f(x)\right\|.$$ Suppose, by contradiction, that $$\lim_{\varepsilon\to0}\sup_{\substack{y\in E\\0<\left\|x-y\right\|_E<\varepsilon}}\frac{|f(x)-f(y)|}{\left\|x-y\right\|_E} > \left\|{\rm D}f(x)\right\|.$$ Then, choosing $\{u_n\}\subset E$ and $\{t_n\}$ such that $\|u_n\|=1$ for all $n$, $t_n\rightarrow 0$ and $$\lim_{\varepsilon\to0}\sup_{\substack{y\in E\\0<\left\|x-y\right\|_E<\varepsilon}}\frac{|f(x)-f(y)|}{\left\|x-y\right\|_E} = \lim_{k\rightarrow \infty} \frac{|f(x+t_n u_n)-f(x)|}{|t_n|},$$ we get $$\|Df(x)(u_n)\| = \dfrac{\|\left( f(x+t_n u_n)- f(x)\right)-\left(-Df(x)(t_n u_n)+ f(x+t_n u_n)-f(x)\right)\|}{|t_n|} \geq $$ $$ \dfrac{\|f(x+t_n u_n)-f(x)\| - \| f(x+t_n u_n)- f(x)-Df(x)(t_n u_n)\|}{|t_n|}.$$ Finelly, for large enough $n$, $$ \|Df(x)(u_n)\|> \|Df(x)\|. $$ A contradiction.