If $f(x+1/n)$ converges to a continuous function $f(x)$ uniformly on $\mathbb{R}$, is $f(x)$ necessarily uniformly continuous? Suppose $f:\mathbb{R}\mapsto\mathbb{R}$ is a continuous function. Define $f_n(x)=f(x+1/n)$ for all $n\in\mathbb{N}^+$. If $f_n(x)\to f(x)$ uniformly on $\mathbb{R}$, can we conclude that $f$ is actually uniformly continuous? If not, can you give a counterexample of $f$ to be not uniformly continuous but satisfies all above conditions?
I know that if $f_n(x)$'s are all uniformly continuous and $f_n(x)\to f(x)$ uniformly, then $f(x)$ must be uniformly continuous. However, here we do not assume $f(x)$ to be uniformly continuous, and $f_n(x)$ has specific structure related to $f(x)$, so at least you cannot directly conclude they are uniformly continuous. I guess the statement above is wrong but it's hard for me to find a counter-example. Can anyone help me?

By assumption, for every $\epsilon >0$, there is an $n_0 \in \Bbb N$ such that for every $x \in \Bbb R$
$$n \geq n_0 \implies \left| f(x) - f \left(x + \frac{1}{n} \right)  \right| < \epsilon$$Now if you repeatedly substitute $x = \pm \frac{1}{n}$ in the above inequality, we can conclude that for any two rationals $a,b$ of the form(say $*$-type) $\frac{m}{n}$, where $m,n \in \Bbb Z, n \geq n_0$, we have
$$|f(a)-f(b)| < \epsilon$$
Now observe that the rationals of the form $m,n \in \Bbb Z, 0<n < n_0$ are nowhere dense in $\Bbb R$. Therefore rationals of $*$-type are dense in $\Bbb R$(we are using the fact that the rationals are dense in $\Bbb R$). Since $f$ is continuous, we obtain that for any two real numbers $x,y$
$$|f(x)-f(y)| \leq \epsilon$$
But since $\epsilon$ was arbitrary, we can conclude that $f$ must be a constant function which is trivially uniformly continuous.
