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Let $f:[0,1] \to \mathbb{R}$ be continuously differentiable. Assume that there is a sequence of continuously differentiable functions $(f_n)_{n=1}^\infty$ such that $$ \lim_{n \to \infty} \|f - f_n\|_{L_\infty} = 0 . $$

Moreover, let $$ \xi := \text{arg}\min_{x \in [0,1]} f(x) \quad \text{ and } \xi_n := \text{arg}\min_{x \in [0,1]} f_n(x) . $$

Is it true that it holds $\xi_n \to \xi$? If yes, why?

Or is it neccessary that the derivatives of $f_n$ converge uniformly to the derivatives of $f$ as well?

Thanks!

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2 Answers 2

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No. Let $f_n(x)=x^n(1-x)$ if $n$ is even, $f_n(x)=x(1-x)^n$ if $n$ is odd. For all $n$ $$ 0\le f_n(x)\le\frac1n,\quad 0\le x\le1. $$ We see that $f_n$ converges uniformly to $0$, but $$ \xi_n=\begin{cases}\dfrac{n}{n+1} & \text{if $n$ is even,}\\ \dfrac{1}{n} & \text{if $n$ is odd.}\end{cases} $$ Thus, $\xi_n$ does not converge.

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The counterexample is wrong, the argmin of $f_n$ is clearly x=0 or x=1 and since the limit function is the zero function all x are the argmin and argmax. The $\xi_n$ are actualy the argmax. Furthermore I think that the original statement is true if the Limit function has a unique argmin but i have no reference.

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