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I learned this questionthis question from math.stackexchange, which is equivalent to ask that if $f:[0,1]\to \mathbb{R}$ is a continuous function with bounded variation, does $$g(x):=\lim_{\epsilon\to 0}\frac{f(x+\epsilon)-f(x-\epsilon)}{2\epsilon}$$ exist for every $x\in[0,1]$ imply that $f$ is absolutely continuous?

Moreover, if we do not know whether $f$ is of bounded variation or not, what can we say about the differentiability of $f$? For example, if $g\equiv 0$, will $f$ be a constant?

Any help is appreciated.

Edit: I think the original question cited from math.stackexchange has been solved by the asker himself there, which is based on the Vitali covering theorem for Radon measures on $\mathbb{R}^n$. For my own question, where $f$ is not assumed to be of bounded variation a priori, Jack Huizenga's answer is good enough for me.

I learned this question from math.stackexchange, which is equivalent to ask that if $f:[0,1]\to \mathbb{R}$ is a continuous function with bounded variation, does $$g(x):=\lim_{\epsilon\to 0}\frac{f(x+\epsilon)-f(x-\epsilon)}{2\epsilon}$$ exist for every $x\in[0,1]$ imply that $f$ is absolutely continuous?

Moreover, if we do not know whether $f$ is of bounded variation or not, what can we say about the differentiability of $f$? For example, if $g\equiv 0$, will $f$ be a constant?

Any help is appreciated.

Edit: I think the original question cited from math.stackexchange has been solved by the asker himself there, which is based on the Vitali covering theorem for Radon measures on $\mathbb{R}^n$. For my own question, where $f$ is not assumed to be of bounded variation a priori, Jack Huizenga's answer is good enough for me.

I learned this question from math.stackexchange, which is equivalent to ask that if $f:[0,1]\to \mathbb{R}$ is a continuous function with bounded variation, does $$g(x):=\lim_{\epsilon\to 0}\frac{f(x+\epsilon)-f(x-\epsilon)}{2\epsilon}$$ exist for every $x\in[0,1]$ imply that $f$ is absolutely continuous?

Moreover, if we do not know whether $f$ is of bounded variation or not, what can we say about the differentiability of $f$? For example, if $g\equiv 0$, will $f$ be a constant?

Any help is appreciated.

Edit: I think the original question cited from math.stackexchange has been solved by the asker himself there, which is based on the Vitali covering theorem for Radon measures on $\mathbb{R}^n$. For my own question, where $f$ is not assumed to be of bounded variation a priori, Jack Huizenga's answer is good enough for me.

deleted 92 characters in body
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qianzhang
qianzhang
  • 183
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I learned this question from math.stackexchange, which is equivalent to ask that if $f:[0,1]\to \mathbb{R}$ is a continuous function with bounded variation, does $$g(x):=\lim_{\epsilon\to 0}\frac{f(x+\epsilon)-f(x-\epsilon)}{2\epsilon}$$ exist for every $x\in[0,1]$ imply that $f$ is absolutely continuous? If the answer is "no", what if $g$ is additionally assumed being integrable or continuous?

Moreover, if we do not know whether $f$ is of bounded variation or not, what can we say about the differentiability of $f$? For example, if $g\equiv 0$, will $f$ be a constant?

Any help is appreciated.

Edit: I think the original question cited from math.stackexchange has been solved by the asker himself there, which is based on the Vitali covering theorem for Radon measures on $\mathbb{R}^n$. For my own question, where $f$ is not assumed to be of bounded variation a priori, Jack Huizenga's answer is good enough for me.

I learned this question from math.stackexchange, which is equivalent to ask that if $f:[0,1]\to \mathbb{R}$ is a continuous function with bounded variation, does $$g(x):=\lim_{\epsilon\to 0}\frac{f(x+\epsilon)-f(x-\epsilon)}{2\epsilon}$$ exist for every $x\in[0,1]$ imply that $f$ is absolutely continuous? If the answer is "no", what if $g$ is additionally assumed being integrable or continuous?

Moreover, if we do not know whether $f$ is of bounded variation or not, what can we say about the differentiability of $f$? For example, if $g\equiv 0$, will $f$ be a constant?

Any help is appreciated.

Edit: I think the original question cited from math.stackexchange has been solved by the asker himself there, which is based on the Vitali covering theorem for Radon measures on $\mathbb{R}^n$. For my own question, where $f$ is not assumed to be of bounded variation a priori, Jack Huizenga's answer is good enough for me.

I learned this question from math.stackexchange, which is equivalent to ask that if $f:[0,1]\to \mathbb{R}$ is a continuous function with bounded variation, does $$g(x):=\lim_{\epsilon\to 0}\frac{f(x+\epsilon)-f(x-\epsilon)}{2\epsilon}$$ exist for every $x\in[0,1]$ imply that $f$ is absolutely continuous?

Moreover, if we do not know whether $f$ is of bounded variation or not, what can we say about the differentiability of $f$? For example, if $g\equiv 0$, will $f$ be a constant?

Any help is appreciated.

Edit: I think the original question cited from math.stackexchange has been solved by the asker himself there, which is based on the Vitali covering theorem for Radon measures on $\mathbb{R}^n$. For my own question, where $f$ is not assumed to be of bounded variation a priori, Jack Huizenga's answer is good enough for me.

added 332 characters in body
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qianzhang
qianzhang
  • 183
  • 6

I learned this question from math.stackchangestackexchange, which is equivalent to ask that if $f:[0,1]\to \mathbb{R}$ is a continuous function with bounded variation, does $$g(x):=\lim_{\epsilon\to 0}\frac{f(x+\epsilon)-f(x-\epsilon)}{2\epsilon}$$ exist for every $x\in[0,1]$ imply that $f$ is absolutely continuous? If the answer is "no", what if $g$ is additionally assumed being integrable or continuous?

Moreover, if we do not know whether $f$ is of bounded variation or not, what can we say about the differentiability of $f$? For example, if $g\equiv 0$, will $f$ be a constant?

Any help is appreciated.

Edit: I think the original question cited from math.stackexchange has been solved by the asker himself there, which is based on the Vitali covering theorem for Radon measures on $\mathbb{R}^n$. For my own question, where $f$ is not assumed to be of bounded variation a priori, Jack Huizenga's answer is good enough for me.

I learned this question from math.stackchange, which is equivalent to ask that if $f:[0,1]\to \mathbb{R}$ is a continuous function with bounded variation, does $$g(x):=\lim_{\epsilon\to 0}\frac{f(x+\epsilon)-f(x-\epsilon)}{2\epsilon}$$ exist for every $x\in[0,1]$ imply that $f$ is absolutely continuous? If the answer is "no", what if $g$ is additionally assumed being integrable or continuous?

Moreover, if we do not know whether $f$ is of bounded variation or not, what can we say about the differentiability of $f$? For example, if $g\equiv 0$, will $f$ be a constant?

Any help is appreciated.

I learned this question from math.stackexchange, which is equivalent to ask that if $f:[0,1]\to \mathbb{R}$ is a continuous function with bounded variation, does $$g(x):=\lim_{\epsilon\to 0}\frac{f(x+\epsilon)-f(x-\epsilon)}{2\epsilon}$$ exist for every $x\in[0,1]$ imply that $f$ is absolutely continuous? If the answer is "no", what if $g$ is additionally assumed being integrable or continuous?

Moreover, if we do not know whether $f$ is of bounded variation or not, what can we say about the differentiability of $f$? For example, if $g\equiv 0$, will $f$ be a constant?

Any help is appreciated.

Edit: I think the original question cited from math.stackexchange has been solved by the asker himself there, which is based on the Vitali covering theorem for Radon measures on $\mathbb{R}^n$. For my own question, where $f$ is not assumed to be of bounded variation a priori, Jack Huizenga's answer is good enough for me.

Source Link
qianzhang
qianzhang
  • 183
  • 6