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Christophe Leuridan
Christophe Leuridan
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The convergence in $L^2$ of the variation ratios does not yield necessarily a pointwise convergence.

For example, consider the case where $f(t,x) = \mathbb{1}_{x = 1/t - \lfloor 1/t \rfloor}$ for all $t>0$ and $x \in [0,1]$, and $f(0,x) = 0$. Each function $f(t,\cdot)$ is null almost everywhere, so $[f(t,\cdot)-f(0,\cdot)]/t \to 0$ in $L^2([0,1])$. Yet, for every $x \in [0,1[$ and $n \in \mathbb{N}$, $f((n+x)^{-1},x)=1$, so the convergences $f(t,x) \to 0$ and $[f(t,\cdot)-f(0,\cdot)]/t \to 0$ do not hold almost surely.

Now, if you have a derivative with regard to $t$ in $L^2$ and a derivative with regard to $t$ at almost every $x \in [0,1]$, those derivatives coincide almost everywhere. This is true because both $L^2$ convergence and almost everywhere convergence imply convergence in measure, for which the limit is almost everywhere unique.

Other possible argument: call $g$ the derivativelimit in $L^2[0,1]$ of $[f(t,\cdot)-f(0,\cdot)]/t$ as $t \to 0$. If you choose a sequence $(t_n)$ tending to $0$ sufficiently fast so that the series $\sum \Vert [f(t_n,\cdot)-f(0,\cdot)]/t_n - g \Vert_2$ converges, then the series $\sum [f(t_n,\cdot)-f(0,\cdot)]/t_n$ converges almost surely (and in $L^2$), so $[f(t_n,\cdot)-f(0,\cdot)]/t_n - g\to 0$ almost surely.

The convergence in $L^2$ of the variation ratios does not yield necessarily a pointwise convergence.

For example, consider the case where $f(t,x) = \mathbb{1}_{x = 1/t - \lfloor 1/t \rfloor}$ for all $t>0$ and $x \in [0,1]$, and $f(0,x) = 0$. Each function $f(t,\cdot)$ is null almost everywhere, so $[f(t,\cdot)-f(0,\cdot)]/t \to 0$ in $L^2([0,1])$. Yet, for every $x \in [0,1[$ and $n \in \mathbb{N}$, $f((n+x)^{-1},x)=1$, so the convergences $f(t,x) \to 0$ and $[f(t,\cdot)-f(0,\cdot)]/t \to 0$ do not hold almost surely.

Now, if you have a derivative with regard to $t$ in $L^2$ and a derivative with regard to $t$ at almost every $x \in [0,1]$, those derivatives coincide almost everywhere. This is true because both $L^2$ convergence and almost everywhere convergence imply convergence in measure, for which the limit is almost everywhere unique.

Other possible argument: call $g$ the derivative in $L^2[0,1]$. If you choose a sequence $(t_n)$ tending to $0$ sufficiently fast so that the series $\sum \Vert [f(t_n,\cdot)-f(0,\cdot)]/t_n - g \Vert_2$ converges, then the series $\sum [f(t_n,\cdot)-f(0,\cdot)]/t_n$ converges almost surely (and in $L^2$), so $[f(t_n,\cdot)-f(0,\cdot)]/t_n - g\to 0$ almost surely.

The convergence in $L^2$ of the variation ratios does not yield necessarily a pointwise convergence.

For example, consider the case where $f(t,x) = \mathbb{1}_{x = 1/t - \lfloor 1/t \rfloor}$ for all $t>0$ and $x \in [0,1]$, and $f(0,x) = 0$. Each function $f(t,\cdot)$ is null almost everywhere, so $[f(t,\cdot)-f(0,\cdot)]/t \to 0$ in $L^2([0,1])$. Yet, for every $x \in [0,1[$ and $n \in \mathbb{N}$, $f((n+x)^{-1},x)=1$, so the convergences $f(t,x) \to 0$ and $[f(t,\cdot)-f(0,\cdot)]/t \to 0$ do not hold almost surely.

Now, if you have a derivative with regard to $t$ in $L^2$ and a derivative with regard to $t$ at almost every $x \in [0,1]$, those derivatives coincide almost everywhere. This is true because both $L^2$ convergence and almost everywhere convergence imply convergence in measure, for which the limit is almost everywhere unique.

Other possible argument: call $g$ the limit in $L^2[0,1]$ of $[f(t,\cdot)-f(0,\cdot)]/t$ as $t \to 0$. If you choose a sequence $(t_n)$ tending to $0$ sufficiently fast so that the series $\sum \Vert [f(t_n,\cdot)-f(0,\cdot)]/t_n - g \Vert_2$ converges, then the series $\sum [f(t_n,\cdot)-f(0,\cdot)]/t_n$ converges almost surely (and in $L^2$), so $[f(t_n,\cdot)-f(0,\cdot)]/t_n - g\to 0$ almost surely.

Source Link
Christophe Leuridan
Christophe Leuridan
  • 3.8k
  • 6
  • 15

The convergence in $L^2$ of the variation ratios does not yield necessarily a pointwise convergence.

For example, consider the case where $f(t,x) = \mathbb{1}_{x = 1/t - \lfloor 1/t \rfloor}$ for all $t>0$ and $x \in [0,1]$, and $f(0,x) = 0$. Each function $f(t,\cdot)$ is null almost everywhere, so $[f(t,\cdot)-f(0,\cdot)]/t \to 0$ in $L^2([0,1])$. Yet, for every $x \in [0,1[$ and $n \in \mathbb{N}$, $f((n+x)^{-1},x)=1$, so the convergences $f(t,x) \to 0$ and $[f(t,\cdot)-f(0,\cdot)]/t \to 0$ do not hold almost surely.

Now, if you have a derivative with regard to $t$ in $L^2$ and a derivative with regard to $t$ at almost every $x \in [0,1]$, those derivatives coincide almost everywhere. This is true because both $L^2$ convergence and almost everywhere convergence imply convergence in measure, for which the limit is almost everywhere unique.

Other possible argument: call $g$ the derivative in $L^2[0,1]$. If you choose a sequence $(t_n)$ tending to $0$ sufficiently fast so that the series $\sum \Vert [f(t_n,\cdot)-f(0,\cdot)]/t_n - g \Vert_2$ converges, then the series $\sum [f(t_n,\cdot)-f(0,\cdot)]/t_n$ converges almost surely (and in $L^2$), so $[f(t_n,\cdot)-f(0,\cdot)]/t_n - g\to 0$ almost surely.