how to prove that$ g_{n}$ converges to$ g$ uniformly on $[0, 1].$

Question

Let $f : R × [0, 1] → R$ be a continuous function and $ {x_{n}}$ a sequence of real numbers converging to $x$. Define

Define $g_{n}(y) = f(x_{n}, y), 0 ≤ y ≤ 1,$

$g(y) = f(x, y), 0 ≤ y ≤ 1.$

Show that$ g_{n}$ converges to$ g$ uniformly on $[0, 1].$

I was trying this $|f_n(x_n,y)-f(x,y)| \leq |f_n(x_n,y)-f_n(x,y)| + |f_n(x,y)-f(x,y)|$

as i don't know how to proceddd further

Pliz help me....

Answer 1

Take $\varepsilon>0$. Note that the set $X=\{x_n\,|\,n\in\mathbb{N}\}\cup\{x\}$ is compact. Therefore, $X\times[0,1]$ is also compact and so the restriction of $f$ to this set is uniformly continuous. Therefore, there is a $\delta>0$ such that, for every $x',x''\in X$ and every $y',y''\in[0,1]$,$$d\bigl((x',y'),(x'',y'')\bigr)<\delta\implies\bigl|f(x',y')-f(x'',y'')\bigr|<\varepsilon.\tag1$$Now, take $N\in\mathbb N$, such that $n\geqslant N\implies|x_n-x|<\delta$. It follows from $(1)$ that$$n\geqslant N\wedge y\in[0,1]\implies\bigl|f(x_n,y)-f(x,y)\bigr|<\varepsilon;$$in other words $(g_n)_{n\in\mathbb N}$ converges uniformly to $g$.

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Note: The compacity of $X$ can be proved as follows: let $\mathcal O$ be an open cover of $X$. Then, since we're dealing with a cover, there is a $A_0\in\mathcal O$ such that $x\in A_0$. Since $A_0$ is open and $\lim_{n\to\infty}x_n=x$, there is a $N\in\mathbb{N}$ such that $n\geqslant N\implies x_n\in A_0$. For each $n\in\{1,2,\ldots,N-1\}$, let $A_n\in\mathcal O$ be such that $x_n\in A_n$. Then$$X\subset\bigcup_{k=0}^{N-1}A_k$$and therefore $\{A_0,A_1,\ldots,A_{n-1}\}$ is a finite subcover of $\mathcal O$.

Answer 2

So the $x$ is fixed, consider the box $I:=[x-1,x+1]\times[0,1]$, so $f$ is uniformly continuous on $I$. Given $\epsilon>0$, there is a $\delta>0$ such that for every $(u,v),(\alpha,\beta)\in I$ and $|u-\alpha|^{2}+|v-\beta|^{2}<\delta$, then $|f(u,v)-f(\alpha,\beta)|<\epsilon$.

Now $x_{n}\rightarrow x$, so some $N$ is such that $|x_{n}-x|<\min\{\delta,1\}$ for all $n\geq N$, then for every $y\in[0,1]$, $|g_{n}(y)-g(y)|=|f(x_{n},y)-f(x,y)|<\epsilon$ for all such $n$.