I'm trying to prove the following result.
Let $(f_n)_{n}$ be a sequence of functions defined on an interval $I$ with real values. We suppose that all $f_n$ are continuous and that the sequence $(f_n)_n$ converges uniformly to a function $f$ on $I$. Then the function $f$ is continuous.
My attempt is the following by using continuity of the $f_n$ and the fact that we have a uniform convergence of the $f_n$ to $f$ which allows to fix an epsilon easily independtly of the $x$ :
Let $\epsilon_1=\frac{\epsilon}{3}$ with $\epsilon>0$. We know that there exists $N\in\mathbb{N}$ such that
$\forall n\geq N,\forall x\in I: \lvert f_n(x)-f(x)\rvert\leq\epsilon_1$
By continuity of functions $f_n$ we know that there exists $\delta>0$ such that $\lvert y-x\rvert\leq\delta$ implies
$\lvert f(y)-f(x)\rvert = \lvert f_n(y)-f_n(y)+f(y)-f(x)+f_n(x)-f_n(x)\rvert\leq\lvert f_n(y)-f(y)\rvert + \lvert f_n(x)-f(x)\rvert + \lvert f_n(y)-f_n(x)\lvert\leq 3\epsilon_1=\epsilon$
for $x,y\in I$
Is this seems correct to you ?
Thank you a lot