Let $U\subset\mathbb{R}^2$ be an open neighborhood of the origin and let $f_n:U\rightarrow \mathbb{R}_{>0}$, $n\in \mathbb{N}$ be a sequence of differentiable functions which uniformly converges on $U$ to an integrable function $f:U\rightarrow \mathbb{R}_{>0}$.

Fix two point $p_1,p_2\in U$ and call $\Gamma= \Gamma_{p_1}^{p_2}$ the set of differentiable paths $\gamma:I\rightarrow U$, $\gamma(t)=(\gamma_1(t),\gamma_2(t))$, such that $\gamma(0)=p_1$, $\gamma(1)=p_2$ and $\gamma(I)\subset U$.

For every $\gamma\in \Gamma$ I'm quite sure it's true: $$\lim_{n\rightarrow \infty}\int_0^1\sqrt{f_n(\gamma(t))(\dot\gamma_1^2+\dot\gamma_2^2)}dt=\int_0^1\sqrt{f(\gamma(t))(\dot\gamma_1^2+\dot\gamma_2^2)}dt,$$ since the functions $f_n$ converge uniformly.

I'm not sure if it's also true $$\lim_{n\rightarrow \infty}\inf_{\gamma\in\Gamma}\int_0^1\sqrt{f_n(\gamma(t))(\dot\gamma_1^2+\dot\gamma_2^2)}dt = \inf_{\gamma\in\Gamma}\int_0^1\sqrt{f(\gamma(t))(\dot\gamma_1^2+\dot\gamma_2^2)}dt $$

Is this equality true or is it only verified under additional hypothesis?


This equality is true. For each $\epsilon>0$ we can choose a path $\gamma$ between $p_1$ and $p_2$ such that $l_g(\gamma)< d_g(p_1,p_2)+\epsilon$. Since $f_n$ converges to $f$ uniformly, we can pick an $N>0$ such that for each $n>N$ we have also $l_{g_n}(\gamma)<l_g(\gamma)+\epsilon$. Therefore $d_{g_n}(p_1,p_2)\leq l_{g_n}(\gamma)<d_g(p_1,p_2)+2\epsilon$. Since $\epsilon$ is arbitary, we obtain $\lim_{n\to\infty} d_{g_n}(p_1,p_2)\leq g(p_1,p_2)$.

Since the path $\gamma$ is compact, $\frac{f_n}{f}$ converges to $1$ uniformly along $\gamma$. Therefore $l_{g_n}(\gamma)$ converges to $l_g(\gamma)$, proving the opposite inequality.

In more detail, given points $p,q$, the sequence $d_{g_n}(p,q)$ converges to some limit $L>0$. For small $\epsilon>0$ negligible compared to $L$ choose $n$ large enough so that $d_{g_n}(p,q)>L-\epsilon$. Then for each curve $\gamma$ between $p$ and $q$ we have $l_{g_n}(\gamma)>L-\epsilon$. In particular if $\gamma$ is a minimizing curve for $g$ we argue as above to get that $L=d_g(p,q)$. If $\gamma$ is not minimizing choose a sufficiently good approximation and argue as before.

  • $\begingroup$ It is not clear to me how to prove the opposite inequality. You have $\ell_{g_n}(\gamma) \to \ell_g(\gamma)$, but the RHS is not $d_g(p_1, p_2)$ . Even if $d_g(p_1, p_2) = \ell_g(\gamma)$, it seems you can at best get again $\le $ but not the opposite one. $\endgroup$ – user99914 Oct 27 '16 at 3:24
  • $\begingroup$ @JohnMa, I tried to clarify this point. Do you have a more elegant argument? $\endgroup$ – Mikhail Katz Oct 27 '16 at 8:18

I want to add more in opposite direction-proof of Mikhail Katz's answer

(1) If $d_g(p,q) >\delta+d_n(p,q)$ for all $n$, there is $N_1$ s.t. $n>N_1\Rightarrow |f_n-f|<\varepsilon_1$

If $c_n$ is a geodesic from $p$ to $q$ wrt $g_n$ then \begin{align*} d_g (p,q) &\leq \int g(c_n',c_n')^\frac{1}{2} \\&\leq \int \sqrt{ g_n(c_n',c_n') +\varepsilon_1 h(c_n',c_n')} \\&\leq d_n(p,q) +\sqrt{\varepsilon_1} {\rm length}_h\ (c_n) \end{align*} where $h=dx^2+dy^2$

Hence $$ \delta <\sqrt{\varepsilon_1} {\rm length}_h\ (c_n) $$

That is $\lim_n\ {\rm length}_h \ (c_n) =\infty$ Then $$ d(p,q) > d_n(p,q)=\int_0^{L_n} \sqrt{f_n(t)} dt $$ where $c_n(t)$ has unit speed wrt $h$ and ${\rm length}_h \ c_n=L_n$

Hence if all $c_n$ is in $B^{h}(p,R)$ for some $R>0$, we have a point $p_n$ in $c_n$ s.t. $f_n(p_n)\rightarrow 0$ This contradicts uniform convergence

So there is $p_n$ in $c_n$ s.t. $d_h(p,p_n)\rightarrow \infty\ \ast$

(2) Consider a sequence of ball : For $q_n\in B_{r_n}^{g_n} (p)$ there is a unique geodesic from $p$ to $q_n$ We have a claim that $\bigcap_n B_{r_n}^{g_n} (p)\supseteq B_r^h(p) $ : If we choose $q\in B_r^h(p) $, then $\ast$ is impossible

Proof of claim : If not there is $q_n$ s.t. $d_h(p,q_n)\rightarrow 0$ and there are two geodesics from $p$ to $q_n$ wrt $g_n$ This means that ${\rm Rm}\ (g_n,s_n)\rightarrow \infty $ for some point $s_n$ around $p$ Since curvature of $g$ around $p$ is bounded above so this is impossible.

  • $\begingroup$ Thanks for the effort. Somehow it would be nice to have a completely transparent proof :-) $\endgroup$ – Mikhail Katz Oct 28 '16 at 8:52
  • $\begingroup$ Thank you for your comment $\endgroup$ – HK Lee Oct 28 '16 at 9:09

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