# real analysis – If \$f_nto f\$ uniformly, then \$min_{[a,b]}f_n to min_{[a,b]}f\$

For $$f_n,f$$ continuous on $$(a,b)$$, and $$f_n to f$$ uniformly, define $$M_n=max_{(a,b)}f_n$$ and $$M=max_{(a,b)}f$$, then it was asked to show that $$M_n to M$$. And it was also asked is it also true for sequence of minima. I did the first part as follows:

$$f_n(x)-f(x)leq max_{(a,b)}lvert f_n(x)-f(x) rvert$$,

$$Rightarrow f_n(x)leq max_{(a,b)}lvert f_n(x)-f(x) rvert + f(x)$$,

Then first taking maximum on right side and then on left side, we will get,

$$M_n-Mleq max_{(a,b)}lvert f_n(x)-f(x) rvert$$.

Similarly, we can also show that,

$$M-M_nleq max_{(a,b)}lvert f_n(x)-f(x) rvert$$, which gives,

$$lvert M_n-M rvert leq max_{(a,b)}lvert f_n(x)-f(x) rvert$$.

By uniform continuity, it is evident that right hand side of above equation goes to zero, which implies $$M_n to M$$.

But I am having difficulty in showing the second part of question. I guess it is true and I thought proof will be similar but the same idea doesn’t seem to work. Any hint. Thanks.