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Problem : If $f$ is continuous on metric space $(X,|\ |)$, then for $\varepsilon >0$ there is locally Lipschitz $f_\varepsilon$ s.t. $$|f(x)- f_\varepsilon (x)|<\varepsilon\ \ast $$

Proof : (1) If $f$ is continuous then it is difference of continuous functions bounded below by $1$ : $$f=({\rm max}\ \{f,0\} +1)-({\rm max}\ \{ -f,0\} +1) $$

(2) Assume that $f\geq 1$ Define $\rho$ on $X$ : $$ |x-y|\leq \rho (x)\Rightarrow |f(x)-f(y)|<\varepsilon $$

Set $$f_\varepsilon (y):=\sup_x \ \bigg\{ f(x)\bigg(1- \frac{|x-y|}{\rho(x)} \bigg) \bigg\} $$ Then $$ f(y)-f_\varepsilon (y) \leq f(y)- \bigg[f(x) - f(x) \frac{|x-y|}{\rho (x) } \bigg]_{x=y} =0 $$

If $$ f_\varepsilon (y) -f (y) =A,\ A:= \sup_x\ B,\ B:= f(x)-f(y) -f(x) |x-y|/\rho(x) $$ we will find an upper bound for $A$ : Then $X =X_1\cup X_2\cup X_3 $ where $$ X_1=\{ x\in X| |x-y|\leq \rho(y) \} $$

$$ X_2=\{x\in X| \rho(y)<|x-y| \leq \rho(x) \} $$

$$ X_3 =\{ x\in X| \rho(y),\ \rho(x) < |x-y| \} $$

Then $${\rm sup}_{x\in X_1 \cup X_2 } B \leq \varepsilon $$

$$ {\rm sup}_{x\in X_3} B \leq -f(y) \leq -1 $$

Here we have $\ast$

But to prove that $f_\varepsilon$ is locally Lipschitz is difficult How can we prove this ? Thank you in anticipation

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  • $\begingroup$ The function $\rho(x)$ does not seem to have any continuity properties. Also, if $f$ is uniformly continuous then you can take $\rho(x)=\delta>0$ to be constant. Then your "approximation" $f_\epsilon(y)$ is not really an approximation of $f$. (Just take $f(x)=|x|+1$ on the real line.) Where did you get this formula? Consider reading instead: projecteuclid.org/download/pdf_1/euclid.rae/1230939175 $\endgroup$ – Moishe Kohan Sep 3 '16 at 12:59
  • $\begingroup$ I missed some detail : $\rho$ is some positive continuous function $\endgroup$ – HK Lee Sep 3 '16 at 13:09
  • $\begingroup$ Reference is 5.6 section 47p. in the following : math.psu.edu/petrunin/papers/alexandrov-geometry/all.pdf $\endgroup$ – HK Lee Sep 3 '16 at 13:12

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