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The definition of uniform continuity is that for any $\epsilon > 0$, we can find specific fixed $\delta>0$ such that for any $x_1,x_2 \in D_f$ we have $|x_1-x_2|<\delta \to |f(x_1)-f(x_2)|<\epsilon$. For any $\epsilon$ we chose, the same $\delta$ must be able to make $|f(x_1)-f(x_2)|<\epsilon$ for any $x_1,x_2 \in D_f$. But I know there's a theorem that says any continuous function is uniformly continuous when its domain is closed. I have a hard time understanding this because a function that is very steep will make $\delta > 0$ to be changed.

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The function is becoming more steep as it gets closer to the right end point, $b$. And clearly, this is function is defined on the closed interval, $[0,b]$. But, we can still see that $\delta$ has to change at different points on the curve to make $|f(x_1)-f(x_2)|<\epsilon$.

What am I missing?

EDIT: http://www.math.uconn.edu/~kconrad/blurbs/analysis/metricspaces.pdf I got my intuition from this paper, on page 30. It shows a graph where one function is continuous and other is uniformly continuous.

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Yes, $\delta$ must be smaller at the right end of your example. So given $\epsilon$, choose a $\delta$ that works at $b$. That $\delta$ will work everywhere.

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    I see it now thank you! Is uniform continuity derived from decreasing nested interval eventually becomes a point? Also, are there some unique uniformly continuous functions other than $\frac{1}{x^k}$ or $x^k$ where $k > 1$? – 2017-01-28
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    You're welcome. I don't understand the followup questions in your comment. As you stated, every continuous function on a closed *bounded* domain is uniformly continuous. None is "unique" in that sense. – 2017-01-28