The connection between the supremums of a function sequence, and the supremum of the limit fumction Suppose we have a function series $\mathbb f_n(x)$ that converge uniformly to $\mathbb f(x)$ for every $x \in [a,b]$ , and suppose that for each $n \in \mathbb N$ and for every $x \in [a,b]$  : $\lvert \mathbb f_n(x)\rvert \le M_n$, i.e $\mathbb f_n(x)$ is bounded for each $n \in \mathbb N$.

I can also prove that $f(x)$ is bounded by some $M_0 \in \Bbb R$.

I am wondering however - does all of the above is enough to prove that the sequence $M_n$ converge to $M_0$?

If the statement is just not true, can someone give me a counter example?
Suppose, by contradiction,that $(M_n)_n$ does not converge to $M_0$. Then for some $r&gt;0$ the set $$S=\{n:|M_0-M_n|&gt;r\}$$ is infinite.  And $\{n:\sup_x|f(x)-f_n(x)|\geq r/2\}$ is finite .So there exists $n\in S$ with $\sup_x |f_n(x)-f(x)|&lt;r/2$.  For such $n$ we have $$(1)\quad M_0\leq M_n+r/2.$$ But there exists $x_n$ with $|f_n(x_n)|&gt;M_n-r/2$. This gives $$M_0\geq |f(x_n)\geq|f_n(x_n)|-|f_n(x_n)-f(x_n)|&gt;   |f(x_n)|-r/2&gt;(M_n-r/2)-r/2=M_n-r$$  from which we have$$(2)\quad M_0\geq M_n -r.$$ From $(1)$ and $(2)$ we have $ M_n+r/2\geq M_0\geq M_n-r,$ so $|M_0-M_n|\leq r,$ a contradiction, because $n\in S.$
Hint : you may prove that in general if $f$ and $g$ are bounded functions then $$| \,||f||_\infty - ||g||_\infty | \leqslant ||f - g||_\infty$$
where as usual $||f||_\infty = \sup |f(x)|$. You can take your inspiration in the proof of the basic inequality $|\, |x|-|y|\,| \leqslant |x-y|$.

Then apply to $g=f_n$.