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Let $g: [0,1]\to \mathbf{R}\cup \{0\}$ be continuous at $0$ with $g(0)=0$. If at $a\in\mathbf{R}$ the function $f:\mathbf{R} \to \mathbf{R}$ satisfies $$\left|f(a+c)-f(a)\right|\leq g\left(|c|\right)$$for all $c\in[-1,1]$ then $f$ is continuous at $a$.

Can somebody please verify that the following proof is rigorous and correct?

We know that for every $\epsilon>0$ there is a $\delta>0$ such that $$|x|<\delta \implies |g(x)|<\epsilon.$$ Since $c\in[-1,1]$ we can replace $x=|c|$ so that $$|c|<\delta \implies|g(|c|)|<\epsilon.$$ We note that as the co-domain of $g$ is the positive reals, $|g(|c|)|=g(|c|)$ and so, using the given inequality $$|c|<\delta \implies |f(a+c)-f(a)|\leq g(|c|)<\epsilon$$ or $$|(a+c)-a|\implies|f(a+c)-f(a)|<\epsilon.$$So $f$ is continuous at $a$.

Thank you.

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Even uniformly continuous in $a$, as your proof correctly shows.

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