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consider an interesting real analysis question:

define average operator on $[0,1]$:

$A_{\epsilon} f (x) = \frac{1}{2\epsilon}\int_{x-\epsilon}^{x+\epsilon} f(y) dy , f \in BV[0,1] $

( may clarify here: $f $ should be regarded as periodic function on $R^1$, i.e. $ f \in BV(S^1) $ )

we know $||A_{\epsilon}f -f||_1 \to 0 $ when $ \epsilon \to 0$.

However its convergence rate is not known in general since the information of $ f $ is not specific, and $ f, \epsilon $ may twist together in convergence rate.

can we find a non trivial convex set $ C \subset BV[0,1] $ s.t. exists $ \gamma>0 $, for all $f \in C $, s.t.

$ ||A_{\epsilon}f -f||_1 \le ||f||_1 \cdot \epsilon^{\gamma} $.

Thanks!

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  • $\begingroup$ What do you mean by trivial? Also, are you willing to accept a uniform implicit constant in the inequality, or does it have to be exactly as you wrote? $\endgroup$
    – Willie Wong
    Commented Aug 17, 2018 at 20:50
  • $\begingroup$ yes, the bound should be the one I wrote. that is why I said convex set. The convex should not be too trivial, like one dim subspace ect. But I am curious what is uniform implicit constant? can you refer me some materials? anything will be helpful. Thanks! $\endgroup$
    – jason
    Commented Aug 18, 2018 at 2:15
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    $\begingroup$ I am asking whether you really want strictly $\| A_\epsilon f - f\|_1 \leq \|f\|_1 \epsilon^\gamma$, or whether you would be happy with a convex set on which $\|A_\epsilon f- f\|_1 \leq 2^{100} \|f\|_1 \epsilon^\gamma$ (where $2^{100}$ is just a placeholder for a suitable constant). In terms of what Luis Silvestre wrote below, it is similarly to the difference between $\|f\|_{W^{\gamma,1}} \leq \|f\|_1$ and, say, $\|f \|_{W^{\gamma,1}} \leq 100 \|f\|_1$. Note that on compact domains by virtue of Poincare inequalities the set $\|f\|_{W^{\gamma,1}} \leq (1+\lambda) \|f\|_1$ is empty for... $\endgroup$
    – Willie Wong
    Commented Aug 18, 2018 at 2:29
  • $\begingroup$ ... all sufficiently small $\lambda$. Alternatively, I've read into your statement an implicit $\forall \epsilon$, but maybe you just want your statement to hold $\forall \epsilon < \epsilon_0$ for some given $\epsilon_0$? $\endgroup$
    – Willie Wong
    Commented Aug 18, 2018 at 2:29
  • $\begingroup$ Note also that your inequality is scaling invariant: if $f$ is in your set, then automatically so is $\lambda f$. Couple this with convexity this means you are automatically looking at a linear subspace. I am not really sure why you prefer to phrase it as a "convex set". $\endgroup$
    – Willie Wong
    Commented Aug 18, 2018 at 2:35

1 Answer 1

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On of the most common expressions for the BV seminorm is $$ [f]_{BV} = \sup_h \frac{\|f(\cdot-h) - f\|_{L^1}}{|h|}.$$

Thus, we have $$ \|A_\varepsilon f - f\| \leq \varepsilon \|f\|_{BV}.$$

I assume that is all that you were looking for. If you really want the $L^1$ norm instead of the BV seminorm on the right hand side, then you need any convex set $C$ that imposes $\|f\|_{BV} \leq C \|f\|_{L^1}$, or perhaps $\|f\|_{W^{\gamma,1}} \lesssim \|f\|_{L^1}$.

By the way $\|A_\varepsilon f - f\|_{L^1} \to 0$ as $\varepsilon \to 0$ for any $f \in L^1$, regardless of whether $f \in BV$.

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  • $\begingroup$ Thanks! yes, I need upper bound should be $||f||_1$. the main problem is to control $ ||f||_{BV, W^{\gamma, 1}} \le ||f||_1$ in certain convex set. $\endgroup$
    – jason
    Commented Aug 18, 2018 at 2:19

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