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} $.


  • $\begingroup$ 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? $\endgroup$ 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
  • 1
    $\begingroup$ 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... $\endgroup$ 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 $\epsilon_0$? $\endgroup$ 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". $\endgroup$ Commented Aug 18, 2018 at 2:35

1 Answer 1


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 $f \in BV$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.