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Let $f_t:H \to H$ for each $t \in [0,T]$ be a map between a Hilbert space $H$. If $f_t(x) \to 0$ as $t \to 0$, and if $x_t \in X$ is such that $x_t \to x$ as $t \to 0$, then under what assumptions can I expect $$f_t(x_t) \to 0$$ as $t \to 0$?

Is uniform covergence of $f_t(x) \to 0$ on compact subsets enough for this?

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    I don't think it is enough, if $f_t(y)=0$ for $y\leq 1/t$ and $f_t(y)=1$ for $y \geq 2/t$ and $x_t = 3/t$ then you get $f_t(x_t)=1$ for every $t$ although $x_t\to 0$ and $f_t(0)=0$ for every $t\in \Bbb N$. – 2017-02-13
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    I guess you need that $f_t$ converge uniformly wrt to the sup norm. – 2017-02-13
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    @Serb do you mean if $\lim_{t \to 0}\sup_{x \in A}f_t(x) = 0$? For all $A \subset H$? Can I get away with $A$ compact? – 2017-02-13
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    well maybe not that strong :). If $A$ is a neighborhood of $x$ and $f_t$ is continuous on $A$ for every $t$ and $\lim_{t\to 0} \sup_{z\in A} \|f_t(z) -f(z)\| = 0$ for some $f$ continuous on $A$ such that $f(x)=0$, then $f_t(x_t)\to 0$ as $t\to 0$. – 2017-02-13
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    It's an argument of the form $||f_t(x_t)|| \leq || f_t(x_t)-f(x_t)||+||f(x_t)-f(x)||$. But maybe there is a possibility to have it under weaker conditions... I dunno – 2017-02-13
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    When you write $f_t(x)\to0$ do you mean that $f_t$ converges to the zero function, or just that $f_t$ converges at $x$ to $0$? – 2017-02-13
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    @s.harp I mean it converges to the zero function. For all $x$ – 2017-02-13

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Uniform convergence on bounded sets is enough:

For if $f_t\to0$ uniformly on bounded sets, you've got for any bounded set $A$ and any $\epsilon$ a value $t'_{(A,\epsilon)}$ so that $t

Now if you are looking at a sequence $x_n$ and $f_{t_n}(x_n)$ with $t_n\to0$, then uniform convergence on compacta is enough. Uniform convergence on compacta is also enough if you have that $t\mapsto x_t$ is continuous in some neighbourhood of $0$.

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    Thanks for replying. So in your first part, you say what I want holds given the uniform convergence on bounded sets. I don't really understand your second part.. under a stronger assumption (uniform boundedness on compact sets) you prove a weaker result, isn't it?? – 2017-02-13
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    @PostName uniform convergence on compact sets is much weaker than uniform convergence on bounded sets. Every compact set is bounded (but not every bounded set is contained in a compact set!). Here we relax the strength of $f_t\to0$ but make the $x_t\to x$ condition stronger so the result stays the same. – 2017-02-13
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    Thanks, one final question. The second part, you are saying if $f_t \to 0$ uniformly on compact sets, and if $x_n \to x$ as $n \to \infty$, then there is sequence $t_n$ with $t_n \to 0$ as $n \to \infty$ such that $f_{t_n}(x_n) \to 0$ as $n \to \infty$. Is that right? I think it is a weaker result that the desired one if I'm correct. – 2017-02-13
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    If $f_t\to0$ uniformly on compacta and $x_n\to x$, then for any sequence $t_n\to0$ you have $f_{t_n}(x_n)\to0$. This result is not comparable with the previous, since while you have a stronger condition on how you are going to the point $x$ via $x_t$, you have a weaker condition on how $f_t\to0$. – 2017-02-13
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    I don't see why $x_n \to x$ is a stronger condition than $x_t \to x$. If $x_t \to x$ then $\hat x_n := x_{(1/n)} \to x$ as $n \to \infty...$ – 2017-02-13
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    Statement A is weaker than B: (A) For every sequence $x_n\to x$ and any $t_n\to0$ we have $f_{t_n}(x_n)\to0$. (B) For every map $t\mapsto x_t$ so that $x_t\to x$ as $t\to0$ we have $f_t(x_t)\to0$ as $t\to0$. A is implied by B but not the other way around. Since the conclusions are the same, this means the condition in A is more of a restriction on what kind of situations can be considered (stronger). – 2017-02-13
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    @s.harp Could you give a hint why continuity of $t \mapsto x_t$ + uniform on compacta would suffice? – 2017-02-13
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    @25Chars Call the map $t\mapsto x_t$ $\gamma$. If $\gamma$ is continuous in some neighbourhood$[0,\kappa)$, then $K:=\gamma([0,\kappa/2])$ is compact since it is the image of a compact set under a continuous mapping. You have $\|f_t\|_K\to0$ and for any $t<\kappa/2$ you have $|f_t(x_t)|≤\|f_t\|_K\to0$. – 2017-02-13