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Consider $\{f_n\}$ a sequence of functions $f_n:[a,b]\times \mathbb{R} \to \mathbb{R}$ continuous in the first variable such that $f_n \to f$.

How can I prove that $$\lim_{n \to \infty}\sup_{\alpha \in [a,b]}f_n(\alpha,x) = \sup_{\alpha \in [a,b]}f(\alpha,x) ?$$

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This statement is clearly false if you don't specify that the convergence $ f_n \rightarrow f $ is uniform, for example take

$ g_n(x) = \begin{cases} nx & \text{if }x \in [0, 1/n] \\ -nx + 2 & \text{if }x \in [1/n, 2/n] \\ 0 & \text{otherwise} \end{cases}$

All $ g_n $ functions are continuous, and $ g_n \rightarrow 0 $, but clearly $ \sup g_n = 1$.


If you suppose uniform convergence, what you actually need to show is that $ f_n(x_n) \rightarrow f(x) $, the supremum inequality for $ f $ being easy to obtain by taking the limit in the ones for $ f_n $.

$$ |f_n(x_n) - f(x)| \leq |f_n(x_n) - f(x_n)| + |f(x_n) - f(x)| \leq \|f_n - f\|_\infty + |f(x_n)-f(x)| \rightarrow 0 $$ gives the needed convergence.

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  • $\begingroup$ alternatively if for fixed $x$ we have that $\|f_n(\cdot,x)-f(\cdot,x)\|_\infty\to 0$ then by the reversed triangle inequality, for finite $\|f_n(\cdot,x)\|_\infty$ and finite $\|f(\cdot,x)\|_\infty$ we have that $$\big|\|f_n(\cdot,x)\|_\infty-\|f(\cdot,x)\|_\infty\big|\le\|f_n(\cdot,x)-f(\cdot,x)\|_\infty$$ $\endgroup$ – Masacroso Apr 15 '17 at 13:34
  • $\begingroup$ What if the convergence is monotonically decreasing instead of uniform? $\endgroup$ – user428573 Apr 15 '17 at 13:38
  • $\begingroup$ @Masacroso but that would only give $ \|f_n\|_\infty \rightarrow \|f\| $, which doesn't provide the result. $\endgroup$ – FreeSalad Apr 15 '17 at 13:39
  • $\begingroup$ @Riku monotonically decreasing in what sense? pointwise monotonical convergence? iirc if you've got $ f_n $ increasing (resp. decreasing) as a function and pointwise increasing (resp. decreasing) convergence of $ f_n $ then you've got uniform convergence $\endgroup$ – FreeSalad Apr 15 '17 at 13:42
  • $\begingroup$ @FreeSalad I dont follow, the statement $\|f_n(\cdot,x)\|_\infty\to\|f(\cdot,x)\|_\infty$ is the title of the question. $\endgroup$ – Masacroso Apr 15 '17 at 13:42

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