last day (23 days later) » 

Maximilian Janisch
Maximilian Janisch
11:22
Hi
@FuzzyPixelz
FuzzyPixelz
FuzzyPixelz
Hello, thank you for joining
I've seen the proof on wikipedia's, it's a bit different from what I'm trying to do
I now realize I needed to show that the function $\phi$ is differentiable before applying to chain rule to it
Maximilian Janisch
Maximilian Janisch
I think you can use the definition of the derivative
FuzzyPixelz
FuzzyPixelz
I mean the function $(x,y,z) \mapsto \int_y^z f(x,t) dt$
It would be complicated to use the definition, no? What about showing that its partial derivatives are continuous?
Maximilian Janisch
Maximilian Janisch
11:37
@FuzzyPixelz I was thinking about something like this:
Leibniz integral rule
In calculus, Leibniz's rule for differentiation under the integral sign, named after Gottfried Leibniz, states that for an integral of the form ∫ a ( x ) b ( x ) f ( x , t ) d t , {\displaystyle \int _{a(x)}^{b(x)}f(x,t)\,\mathrm {d} t,} where − ∞ <...
This proves that $\phi$ is differentiable w.r.t. $x$ (and has continuous partial derivative)
Proving that $\phi$ is continuously differentiable w.r.t. $y$ and $z$ seems to be trivial with the help of the Fundamental Theorem of calculus
@FuzzyPixelz All clear?
FuzzyPixelz
FuzzyPixelz
11:57
@MaximilianJanisch I don't see how it proves that the partial derivative w.r.t $x$ is continuous
Maximilian Janisch
Maximilian Janisch
@FuzzyPixelz But $\frac{\partial f}{\partial x}$ is continuous
by assumption
by the way everyone we are talking about math.stackexchange.com/questions/3572368
So try changing $x$ a little bit to see that $\frac{\partial\phi}{\partial x}$ (which is given explicitly in terms of $\frac{\partial f}{\partial x}$) is also continuous
All clear now @FuzzyPixelz ?
FuzzyPixelz
FuzzyPixelz
I'm sorry, but since the variable here is $(x,y,z)$ it wouldn't be enough to say that $\frac{\partial f}{\partial x}$ is continuous
Maximilian Janisch
Maximilian Janisch
Well ok see what happens if you change $x$, $y$ and $z$ a little bit, say by $\varepsilon$
(Since all norms on $\mathbb R^3$ are equivalent you can choose the sup norm for convenience)
FuzzyPixelz
FuzzyPixelz
Yes, in general continuity w.r.t to each variable doesn't imply continuity, right?
Maximilian Janisch
Maximilian Janisch
Indeed, the standard example is $g(x,y)= x y/(x^2+y^2)$ if $x,y$ are not both $0$ and $g(0,0)=0$ if $x=y=0$
This is not continuous at (0,0) but continuous w.r.t. x and w.r.t. y separately
@FuzzyPixelz
I have to go now by the way
see you
FuzzyPixelz
FuzzyPixelz
12:12
Okay, thanks for everything @MaximilianJanisch
FuzzyPixelz
FuzzyPixelz
13:07
@MaximilianJanisch I figured it out, so I wrote an answer, math.stackexchange.com/questions/3572368/…
 
10 hours later…
Maximilian Janisch
Maximilian Janisch
22:59
@FuzzyPixelz There is a small mistake kn the proof that needs to be fixed (see my comment there). In essence you have to use continuity of $\frac{\mathrm df}{\mathrm dx}$ to say that changing $x$ doesn’t change the overall integral that much

  last day (23 days later) »