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Let $T\colon X \to Y$ be a nonlinear operator between Hilbert spaces which is Lipschitz and is Hadamard differentiable. It satisfies $$T(x+th)=T(x) + tT'(x)(h) + r(t)$$ where $r(t)=r(t,x,h)$ is the remainder term which is such that $$\frac{r(t,x,h)}{t} \to 0\text{ in $Y$ as $t \to 0$}.$$ Since $T$ is Hadamard, this convergence is uniform in $h$ on compact subsets. Is there any condition under which we can say that this convergence is also uniform with respect to $x$ whenever $x$ belongs to a bounded/compact/something-else subset?

If it helps, $X$ is compactly embedded in $Y$.

I am asking since I'm interested particularly in the above limit when I have instead of $x$ a sequence $x_n$ (such that $x_n \to x$) and I desire a uniform in $n$ convergence as $t \to 0$.


$T$ is Hadamard differentiable at $x$ in the direction $h$ if for all sequences $h_n \to h$ and all non-negative sequences $t_n \to 0$, we have the existence of the limit $$\lim_{n \to \infty}\frac{T(x+t_nh_n)-T(x)}{t_n}$$ and the limit then equals $T'(x)(h)$.

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    $\begingroup$ Can you recall the definition of the Hadamard derivative? $\endgroup$ Commented Jun 20, 2018 at 13:33
  • $\begingroup$ @PiotrHajlasz I edited $\endgroup$
    – M.L
    Commented Jun 20, 2018 at 13:53
  • $\begingroup$ Regarding the definition of the Hadamard derivative. What is $T'(x)$? Is it directional derivative? Do you assume linearity in $h$? $\endgroup$ Commented Jun 20, 2018 at 16:23
  • $\begingroup$ Yes, $T'(x)h$ is the directional derivative at the point $x$ in the direction $h$. Linearity is not assumed. $\endgroup$
    – M.L
    Commented Jun 20, 2018 at 17:03

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