On uniform convergence

Question

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

$g_n(y)=f(x_n,y)$, $0\leq y\leq1$,

$g(y)=f(x,y)$, $0\leq y\leq1$.

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!

Answer 1

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 $y\in\left[0,1\right]$, $\exists N_{\epsilon,y}\in\mathbb{N}$ such that $\left|f(x_n,y)-f(x,y)\right|<\epsilon$ whenever $n\geq N_{\epsilon,y}$.

Now, $\left[0,1\right]=\bigcup_{y\in\left[0,1\right]}(y-\epsilon,y+\epsilon)$. As $\left[0,1\right]$ is compact, there exists a finite subcover. Thus, there exists $m\in\mathbb{N}$ such that $\left[0,1\right]=\bigcup\limits_{i=1}^{m}(y_i-\epsilon,y_i+\epsilon)$. Now any $y\in\left[0,1\right]$ 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 $\left[0,1\right]$.

Answer 2

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:

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

Answer 3

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.

Answer 4

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

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