Existence of Partial Derivatives and Continuity ....

[Math Amateur]

I am reading the book "Several Real Variables" by Shmuel Kantorovitz ... ...

I am currently focused on Chapter 2: Derivation ... ...

I need help with another element of the proof of Kantorovitz's Proposition on pages 61-62 ...

Kantorovitz's Proposition on pages 61-62 reads as follows:

In the above proof we read the following:

" ... ... Formula 2.4 is trivially true in case $h_j = 0$, and by (2.2) - (2.4)

$f(x + h) - f(x) = \sum_{ j = 1}^k [ F_j ( h_j ) - F_j (0) ]$

$= \sum_j h_j \frac{ \partial f }{ \partial x_j } ( x + h^{ j - 1 } + \theta_j h_j e^j )$ ... ... ... ... ... "I have tried to derive $f(x + h) - f(x) = \sum_{ j = 1}^k [ F_j ( h_j ) - F_j (0) ]$ but did not succeed ...

... can someone please show how $f(x + h) - f(x)$ equals $\sum_{ j = 1}^k [ F_j ( h_j ) - F_j (0) ]$ ...Also can someone show how the above equals $\sum_j h_j \frac{ \partial f }{ \partial x_j } ( x + h^{ j - 1 } + \theta_j h_j e^j )$ ... ...

Help will be much appreciated ... ...

Peter

 

[andrewkirk]

Hello Peter. It's good to see you posting again. Haven't seen you for some time.

I can't make sense of the line that defines $F_j(t)$, between 2.2 and 2.3. There's a big space containing a fuzzy superscript-like mark that is a bit like a $t$ or a 1, but doesn't exactly match either. I can't think of any way to interpret it. It can't be an exponent, as a vector can't be raised to a power.

Do you know what that line is trying to do?

PS Also $e^j$ appears undefined. Is the author referring to the vector with all zero components except for a 1 in the $j$th position?

 

[andrewkirk]

Ah OK, got it now. It's actually $h^{j-1}$ but the minus sign is invisible in the scan, and the 1 doesn't quite look like a 1.

To prove that formula, look back at 2.2. All we need to do is prove that $F_j(h_j)=f(x+h^j)$ and $F_j(0)=f(x+h^{j-1})$. To do that, just substitute $h_j$ and 0 into the formula that defines $F_j$. It's easy for the 0 case. The other case might require a bit of extra thought, involving the relationship between $h^j$ and $h^{j-1}$.

For the last bit, substitute the RHS of 2.4 into the bit inside the absolute value signs in the next line. Then use 2.3 to replace the derivative of $F_j$ by a partial derivative of $f$ (RHS of 2.3), inside that absolute value.

But how to get rid of the absolute value signs? I think the answer is that they should not be there. They are a mistake. Look at where they are introduced in 2.2. No reason is given for them and inspection of that formula suggests it makes more sense if the absolute value signs are just replaced by parentheses. Indeed, for 2.2 to be true as written, $x$ would have to be a local minimum of $f$, and that is not in the premises of the proposition. The absolute-valueness of them is not used anywhere in the proof. Indeed, the proof goes through if they are changed to parentheses, but not otherwise.

 

[Math Amateur]

Ah OK, got it now. It's actually $h^{j-1}$ but the minus sign is invisible in the scan, and the 1 doesn't quite look like a 1.

To prove that formula, look back at 2.2. All we need to do is prove that $F_j(h_j)=f(x+h^j)$ and $F_j(0)=f(x+h^{j-1})$. To do that, just substitute $h_j$ and 0 into the formula that defines $F_j$. It's easy for the 0 case. The other case might require a bit of extra thought, involving the relationship between $h^j$ and $h^{j-1}$.

For the last bit, substitute the RHS of 2.4 into the bit inside the absolute value signs in the next line. Then use 2.3 to replace the derivative of $F_j$ by a partial derivative of $f$ (RHS of 2.3), inside that absolute value.

But how to get rid of the absolute value signs? I think the answer is that they should not be there. They are a mistake. Look at where they are introduced in 2.2. No reason is given for them and inspection of that formula suggests it makes more sense if the absolute value signs are just replaced by parentheses. Indeed, for 2.2 to be true as written, $x$ would have to be a local minimum of $f$, and that is not in the premises of the proposition. The absolute-valueness of them is not used anywhere in the proof. Indeed, the proof goes through if they are changed to parentheses, but not otherwise.

Hi Andrew,

Good to hear from you ...

Sorry about difficulty of reading scan ... it's to do with the nature and quality of the printing in the text ... in particular ... brackets ... ie [ and ] look like absolute value signs ... sorry about misleading you ...

Will now read your next post ...

Peter

 

[Math Amateur]

Thanks so much, Andrew ...

Did what you suggested ... and result was achieved ... !

Thanks again!

Peter