Let $f:\mathbb{R}\rightarrow [0,+\infty)$ continuous and let $a_n\in \mathbb{R}$ a sequence with $a_n\to 0$. Which of the following statements are correct?
(a) If $f_n(x)=f(x+a_n)$ then $f_n\to f$ uniformly.
(b) (f $f_n(x)=f(x)^{1+a_n}$ then $f_n\to f$ uniformly.
(c) If $f_n(x)=f(x)+a_n$ then $f_n\to f$ uniformly.
(d) If $f_n(x)=(1+a_n)f(x)$ then $f_n\to f$ uniformly.
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I have done the following :
At (a) we have $|f_n(x)-f(x)|=|f(x+a_n)-f(x)|\leq \epsilon$ with $|x+a_n-x|=|a_n|\leq \delta$ from continuity. So this is true.
(b) is false.
At (c) we have $|f_n(x)-f(x)|=|a_n|\leq \epsilon$ since $a_n\to 0$. So this is true.
At (c) we have $|f_n(x)-f(x)|=|a_nf(x)| $. This is false.
Is everything correct?
a) is false. You proved pointwise convergence and concluded uniform convergence. For a counter-example let $f(x)=x^{2}$ and $a_n=\frac 1 n$. Uniform convergence would give $|(x+\frac 1 n)^{2}-x^{2}|<\epsilon$ for all $x$ as long as $n >N$ with $N$ independent of $x$. Obviously the left side can be made as large as you want by choosing $x$ large so the convergence is not uniform.
Your answer for c) is right. For b) and d) you have to give specific examples to say that they are false. In both cases you can take $f(x)=x$ and $a_n=\frac 1n$