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Let $f$ and $g$ be two continuous functions on an interval $[a,b] \subset \mathbb{R}$. Define for every $k \in \mathbb{N}$

$f^k(x)=(1-\varepsilon_k)f(x)+\varepsilon_kg(x)$

in which $\varepsilon_k \rightarrow 0$ as $k \rightarrow \infty$. Suppose that function $f$ has a unique maximizer $x=y$. How can I show that there is a sequence of $\{y^k\}_{k=1}^{\infty}$ in which $y^k$ is a maximizer of $f^k$ for every $k$, such that $y^k \rightarrow y$ when $k \rightarrow \infty$?

elnaz
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1 Answers1

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We have $f^k$ converges to $f$ uniformly. Take $y^k$ to be any maximizer of $f^k$ which exists since $[a,b]$ is compact. Shall show $y^k$ converges to $y$. Consider any convergent subsequence of $y^k$. Then we have $f^k(y^k)≥f^k(x) \forall x\in [a,b]$ . Now letting $k$ go to infinity we have since $ f^k $ converges uniformly and $y_{k_l}$ converges to $y'$ say we have $f^{k_l}(y_{k_l})$ converges to $f(y')$ and $f^{k_l}(x)$ converges to $f(x)$. So we $f(y')≥f(x) \forall x \in [a,b]$. So by uniqueness of maximizer $y'=y$. Thus any convergent subsequence has the same limit $y$. So $y^k$ converges to $y$.

  • Which book do you follow in functional analysis? –  May 07 '17 at 14:05
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    Rudin is one of those. – elnaz May 07 '17 at 14:07
  • I was wondering about the last step of your proof, where you conclude that $y^k$ converges to $y$. How can you derive that? As far as i understand, your proof only shows that any convergent sub-sequence of $y^k$ is going to converge to y. – elnaz May 10 '17 at 14:34
  • A sequence of bounded real numbers say $y_n$ is convergent iff every convergent subsequence of this sequence converges to the same limit. –  May 10 '17 at 14:43