I can't figure out how to prove this statement.
If $I$ is a bounded interval and $f$ is an unbounded function defined on $I$, then $f$ cannot be uniformly continuous.
It seems pretty obvious to me that this statement is true, and I figure it has to do with the fact that $|x-x_0|$ is, at most, equal to $|b-a|$ if $I$ is defined to be $(a,b)$ while $|f(x)-f(x_0)|$ could be infinite. Help?
Best Answer
First, $|f(x)-f(x_0)|$ cannot be infinite if $f$ is a real valued function. You just cannot bound it by a constant for all $x$ simultaneously.
Secondly you just can stick to the defitions. If $f$ where UC, then to, say, $\varepsilon = 1$ there would be $\delta > 0$ st $|x-y| < \delta \Rightarrow |f(x)-f(y)| < 1$. Now choose $n $ such that $n\delta > |b-a| $ (assuming $I=(a,b)$)
Using the triangle inequality you can then easily derive that for any $x,x_0$ in $(a, b)$, $|f(x) - f(x_0)|\le n\cdot 1 = n$ , so $f$ is bounded, contradicting the assumption.
Assume, e.g.,that $x< x_0$
The last inequality is then true since you can find $k$ points $\{x_i\}$ such that $k\le n$ and $x< x_1 <\ldots x_k < x_0$ such that the difference between $x, x_1$, between $x_k, x_0$ and between each pair $x_i, x_{i+1}$ is less than $\delta$, so $$|f(x)-f(x_0)| = |f(x)-f(x_1) +f(x_1) -\ldots - f(x_k)+f(x_k) -f(x_0)|\le |f(x)-f(x_1)|+|f(x_1)-f(x_2)|+\ldots+|f(x_k)-f(x_0)|\le k $$