# Showing $f$ is continuous given $||f(x)-f(y)|| ≤ u(||x-y||)$ where $\lim_{t \to 0+} u(t)=0$

Let $u: [0; +∞[→R$ be a function so that $\lim_{t \to 0+} u(t)=0$, $A$ be a subset of $R^n$ and $f: A → R^m$ be a transformation that satisfies the condition $$||f(x)-f(y)|| ≤ u(||x-y||)$$ for all $x$ and $y$ in $A$. Show that $f$ is continuous in $A$ (i.e. continuous in every point in $A$).

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What have you tried? Stronger and no more difficult, you can show that $f$ is uniformly continuous. –  Jonas Meyer Jan 28 '13 at 5:46

Note that the value $u(0)$ is immaterial with respect to the continuity properties of $f$. With this in mind, let $\hat{u} = u \cdot 1_{(0,\infty)}$, and note that the condition on $u$ implies that $\hat{u}$ is continuous at $0$, and of course, we still have $\|f(x)-f(y)\| \leq \hat{u}(\|x-y\|)$.

Hint: Now let $\epsilon>0$. Then there exists a $\delta>0$ such that if $t \in [0,\delta)$ then $\hat{u}(t) \in [0,\epsilon)$.

Mouse over for a much stronger hint:

Suppose $\|x-y\| < \delta$. Then the above shows that $\hat{u}(\|x-y\|) < \epsilon$.

You should be able to finish it from here.

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$$||f(x+h)-f(x)|| ≤ u(||(x+h)-x||)= u(||h||) .$$
Now, notice this, as $||h|| \to 0$, we have $u(||h||) \to 0$ by the assumption $\lim_{ t \to 0^{+} } u(t)=0.$