Full problem statement: Let $X \subset \mathbb{R}^m$ be compact and $f : X \rightarrow \mathbb{R}$ be continuous. Given $\epsilon > 0$, show that there is a constant $M$ such that for all $x,y \in X$ we have $|f(x) - f(y)| \leq M |x-y| + \epsilon$.
Please check my solution for correctness below:
Solution: Assume on the contrary that there exists an $\epsilon > 0$ such that $\forall M \ \exists x,y \in X$ s.t. $|f(x) - f(y)| > M|x-y| + \epsilon$. First note that $Im\ f = f(X) \subset \mathbb{R}$ is the continuous image of a compact set, so it is compact, and so, closed and bounded. By noting that the left hand side is bounded above and by taking large enough values of $M$, we see that there are two sequences of points in $X$- $(x_n)_0^\infty, (y_n)_0^\infty$ such that $d_X(x_n, y_n) < \frac{1}{2^n}$ while $|f(x_n) - f(y_n)| > \epsilon$ for all $n \in \mathbb{N}_0$. By compactness of $X$, $(x_n)$ has a subsequence $x_{n_k}$ that converges to a point $l_1 \in X$. Again, $(y_{n_k})$ has subsequence $(y_{n_{k(l)}})$ that converges to a limit $l_2 \in X$. Considering the fact that $d_X(x_{n_{k(l)}}, y_{n_{k(l)}}) < \frac{1}{2^n}$, we conclude that $l_1 = l_2$. That is, we have, $(x_{n_{k(l)}}) \rightarrow l,\ (y_{n_{k(l)}}) \rightarrow l$ for some $l \in X$ (compactness). By continuity of $f$, we must have the sequences $f(x_{n_{k(l)}})$ and $f(y_{n_{k(l)}})$ converge to the same limit $f(l) \in \mathbb{R}$. But this is not possible because $|f(x_{n_{k(l)}}) - f(y_{n_{k(l)}})| > \epsilon$ for all $n_{k(l)}$. Thus, there cannot exist such an $\epsilon > 0$, and the proposition follows. $\square$
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