calculus - Convergence of length under uniform convergence of closed curves?

Suppose $\gamma:\mathbb{S}^1\rightarrow\mathbb{R}^2$ is a finite length closed curve and $\{\gamma_t\}_{t>0}$ is a set of smooth closed curves with the following property: for all $\epsilon$, we can find a time $T$ such that for every $t\in(0,T)$, there exists a parametrisation of $\gamma_t$ such that
\begin{equation}
\sup_{x\in\mathbb{S}^1} d(\gamma(x),\gamma_t(x))<\epsilon
\end{equation}
Then is it true that the length of $\gamma_t$ converges to the length of $\gamma$ as $t\rightarrow0$? My intuitive interpretation of the property is that $\gamma_t$ converges to $\gamma$ with no 'overlaps' (which may be allowed by a weaker notion of convergence such as Hausdorff convergence), so maybe the lengths do converge.

But then the following idea might be (related to?) a counterexample: take the image of $\gamma$ to just be a unit circle, and consider $\gamma_n(x)=\sin(nx)/n$ on $[0,2\pi]$ with the endpoints identified, so we get some sort of $\sin$ graph over the unit circle. Then I think this converges uniformly to the unit circle as $n\rightarrow\infty$ but the lengths don't converge: in fact the length of such a curve is
\begin{equation}
\int_{0}^{2\pi}\sqrt{(1+cos^2(nx))}\,dx\approx7.6404
\end{equation}

independently of $n$, according to Wolfram alpha. This is not equal to $2\pi$ and hence the lengths don't converge to that of the unit circle. I'm not sure how to translate this into a counterexample for $t\in\mathbb{R}_{>0}$ rather than integers $n\in\mathbb{N}$ (for which taking the interval $[0,2\pi]$ always works). Any help would be appreciated!

EDIT: let's suppose $\gamma$ is a Jordan curve to prevent a degenerate case where $\gamma$ is a point, for example.

Answer

It is not true. Here is a counterexample: Let $\gamma: [0,2\pi]\rightarrow \mathbb{R}^2, \ \gamma(x)=(0,0)$ and $\gamma_t: [0,2\pi]\rightarrow \mathbb{R}^2, \ \gamma_t(x)=t \left(\cos(x/t^2), \ \sin(x/t^2)\right)$. This family of smooth curves does clearly satisfy your conditions (converging in the sup-norm to $\gamma$). However, we have

$$ L(\gamma)=0 $$

but

$$ L(\gamma_t)= \int_0^{2\pi} \vert \gamma_t '(x)\vert dx = \int_0^{2\pi} \frac{1}{t} dx = \frac{2\pi}{t}.$$

Edit:

You can use your own construction to give a counterexample. Namely, let for $n\in \mathbb{N}$ $\gamma_n$ be as you defined. Then the you get a counterexample by

$$ \tau_t = \begin{cases} \gamma_n,& t=1/n; \\ \gamma,& else. \end{cases} $$

Then the limit $ L(\tau_t)$ does not converge to $L(\gamma)$ (it does not converge at all).

If you only want $\gamma$ be a Jordan curve, then you use my curves above to construct a counterexample. Let $\gamma$ be a Jordan curve. Then we set
$$\gamma_t: [0,2\pi]\rightarrow \mathbb{R}^2, \ \gamma_t(x)=\gamma(x) + t (\cos(x/t^2), \ \sin(x/t^2))$$

We have

$$ L(\gamma_t) = \int_0^{2\pi} \vert \gamma'(x) + \gamma_t ' (x) \vert dx \geq \int_0^{2\pi} \vert \gamma_t '(x)\vert dx - \int_0^{2\pi} \vert \gamma'(x)\vert dx = \frac{2\pi}{t} - L(\gamma).$$

Hence, $\lim_{t\downarrow 0} L(\gamma_t) = +\infty.$