Dtjytyk

Let $f:mathbb{R}times [0,1]tomathbb{R}$ be a continuous function and ${x_n}$ a sequence of real numbers converging to $x$. Define

$g_n(y)=f(x_n,y)$, $0leq yleq1$,

$g(y)=f(x,y)$, $0leq yleq1$.

Show that $g_n$ converges to $g$ uniformly on $[0,1]$.

I don't need a complete solution, but a bit of direction would be extremely useful. Thank you!

Let $K={x,x_1,x_2,cdots}$. Then $K$ is a compact set and $f$ is continuous, hence uniformly continuous on $K times [0,1]$. Now just write down the definition of uniform continuity and you will get the conclusion.

Credit to Kavi.

$f$ is uniformly continuos on $K={x,x_1,x_2,...}×[0,y],$ compact.

$epsilon >0$ given, there exists a $delta >0$ s.t.

$||(x,'y')-(x,y)|| lt delta$ implies

$|f(x',y')-f(x,y)| lt epsilon$.

With $y'=y$:

$||(x',y)-(x,y)|| lt delta$ implies

$|f(x',y)-f(x,y)| lt epsilon$, i.e.

$|x'-x| lt delta$ implies

$|f(x',y)-f(x,y)| lt epsilon$.

Since $x_n$ converges to $x$:

For a $delta >0$ there is a $n_0$ s.t.

for $n ge n_0$ :

We have $|x_n -x| lt delta$ which implies

$|f(x_n,y)-f(x,y)| lt epsilon$.

Let $epsilon>0$ be given.

Now, $left|g_n(y)-g(y)right|=left|f(x_n,y)-f(x,y)right|$

As $f$ is continuous, therefore for each $yinleft[0,1right]$, $exists N_{epsilon,y}inmathbb{N}$ such that $left|f(x_n,y)-f(x,y)right|<epsilon$ whenever $ngeq N_{epsilon,y}$.

Now, $left[0,1right]=bigcup_{yinleft[0,1right]}(y-epsilon,y+epsilon)$. As $left[0,1right]$ is compact, there exists a finite subcover. Thus, there exists $minmathbb{N}$ such that $left[0,1right]=bigcuplimits_{i=1}^{m}(y_i-epsilon,y_i+epsilon)$. Now any $yinleft[0,1right]$ belongs to one of the intervals $(y_i-epsilon,y_i+epsilon)$ for some $i$. Thus, taking $N=$ max${N_{epsilon,y_i}: i=1,2,...,m}$, we get that $g_n$ converges to $g$ uniformly on $left[0,1right]$.

Here's a certainly much cleaner way (tbh, I didn't check your solution properly):

We want to show that $g_n$ converges uniformly to $g$, which is the same as asking that $$lim_{n to + infty} sup_{y in [0,1]} vert g_n(y)-g(y) vert =0.$$ Now observe that $vert g_n(y)-g(y)vert$ is a continuous function on $[0,1]$ for each $n$, hence it attains its maximum. Therefore denote $y_n in [0,1]$ such that $$sup_{y in [0,1]} vert g_n(y)-g(y) vert = vert g_n(y_n)-g(y_n) vert.$$ At the same time we have $$vert g_n(y_n)-g(y_n) vert=vert f(x_n,y_n)-f(x,y_n) vert.$$ Since $[0,1]$ is compact, we can pick a subsequence $y_{n_k}$ such that $y_{n_k}$ converges to some $y$ and as $x_n$ already converges to $x$ and so will its corresponding subsequence $x_{n_k}$.

Using that $f$ is continuous in both entries now gives the desired claim:

$$lim_{n to infty}sup_{y in [0,1]}vert g_n(y)-g(y) vert=lim_{k to infty} vert f(x_{n_k},y_{n_k})-f(x,y_{n_k})vert=0. $$