# Uniformly continuous functions sequence $f_n(x)$ converges uniformly to a uniformly continuous function $f(x)$? [duplicate]

We know that if continuous functions sequence $g_n(x)$ converges uniformly to $g(x)$, then $g(x)$ is continuous function.

But what if uniformly continuous functions sequence $f_n(x)$ converges uniformly to $f(x)$? Does that imply $f(x)$ is uniformly continuous function?

If so - how do we prove it? what $\delta$ should we take for $f(x)$? We can't simply choose $\min\{ \delta_n \}$.

And what if $f_n(x)$ converges pointwise to $f(x)$. Will $f(x)$ be continuous? uniformly continuous? Or pointwise convergence is not enough?

## marked as duplicate by user99914, Claude Leibovici calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Feb 20 '16 at 9:02

Claim: Let $f_n$ be a sequence of uniformly continuous functions which converges uniformly to $f$, then $f$ is uniformly continuous.
Proof: Let $\epsilon >0$. Since $f_n \to f$ uniformly, there is $f_n$ so that $$|f_n(x) - f(x) | < \epsilon /3$$ for all $x$. Since $f_n$ is uniformly continuous, there is $\delta >0$ so that $$|f_n(x) - f_n(y)|<\epsilon/3$$ whenever $|x-y|<\delta$. Thus
$$|f(x) - f(y)| \le |f(x) - f_n(x) | + |f_n(x) - f_n(y)| + |f_n(y) - f(y)| < \epsilon$$ whenever $|x-y|<\delta$. Thus $f$ is uniformly continuous.
If $f_n \to f$ only pointwisely, $f$ might not even be continuous (think of $x^n$ on $[0,1]$)