Vtyjkyuk

Let $Asubset mathbbR$ and suppose that $f:Ato mathbbR$ has the following property: for each $epsilon>0$ there exists a function $g_epsilon:Ato mathbbR$ such that $g_epsilon$ is uniformly continuous on $A$ and $|f(x)-g_epsilon(x)|<epsilon$ for all $xin A$. Prove that $f$ is uniformly continuous on A.

My attempt: Since $g_epsilon$ is uniformly continuous so for each $epsilon>0$ $exists delta(epsilon)>0$ such that if $x,uin A$ are arbitrary numbers then $|g_epsilon/2(x)-f(x)|<epsilon/2$. Now

$|f(x)-f(u)|=|f(x)-g_epsilon/2(x)+g_epsilon/2(x)-f(u)|leq |f(x)-g_epsilon/2(x)|+|g_epsilon/2(x)-f(u)|<epsilon/2+epsilon/2=epsilon$. Hence $f(x)$ is uniformly continuous.

Is my solution correct? If not, where is my mistake? Please any suggestion would be helpful. thank you.

Your idea is great. But you can't compare $f(u)$ and $g_epsilon/2(x)$ that easily, so you need to include one more term. Also, you have to use your $delta$ somehow, and as part of that specify what property you want it to fulfill. At the moment, you've just picked an arbitrary $delta$ and not used it.

Take a $delta$ such that for any $x, y$ with $|x-y|<delta$ we have $|g_epsilon/3(x)-g_epsilon/3(y)|<fracepsilon3$. This way we get, for any $x, y$ with $|x, y|<delta$:
$$
|f(x)-f(y)| = |f(x)-g_epsilon/3(x) + g_epsilon/3(x) - g_epsilon/3(u) + g_epsilon/3(u) - f(u)|\
leq |f(x)-g_epsilon/3(x)| + |g_epsilon/3(x) - g_epsilon/3(u)| + |g_epsilon/3(u) - f(u)|\
<3cdot fracepsilon3
$$

One term missing.
$$
|f(x)- f(y)| leqslant |f(x) - g_varepsilon/3(x)| + |g_varepsilon/3(x) - g_varepsilon/3(y)|+|g_varepsilon/3(y) - f(y)|.
$$