# Limit and sup: $\lim_{n \to \infty}\sup_{\alpha \in [a,b]}f_n(\alpha,x) = \sup_{\alpha \in [a,b]}f(\alpha,x)$

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

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.

• 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 $$\big|\|f_n(\cdot,x)\|_\infty-\|f(\cdot,x)\|_\infty\big|\le\|f_n(\cdot,x)-f(\cdot,x)\|_\infty$$ – Masacroso Apr 15 '17 at 13:34
• What if the convergence is monotonically decreasing instead of uniform? – user428573 Apr 15 '17 at 13:38
• @Masacroso but that would only give $\|f_n\|_\infty \rightarrow \|f\|$, which doesn't provide the result. – FreeSalad Apr 15 '17 at 13:39
• @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 – FreeSalad Apr 15 '17 at 13:42
• @FreeSalad I dont follow, the statement $\|f_n(\cdot,x)\|_\infty\to\|f(\cdot,x)\|_\infty$ is the title of the question. – Masacroso Apr 15 '17 at 13:42