# Confusion with an inequality involving norms

1. Sep 9, 2012

### naaa00

1. The problem statement, all variables and given/known data

Hello,

I'm little bit confused about a particular inequality in a proof:

| (D_j f_i) (y) - (D_j f_i) (x) | ≤ | [(f'(y) - f'(x)]e_j | ≤ ||f'(y) - f'(x)||

The last part of the inequality confuses me. Is the absolute value (norm on R) less than any other norm on R^n?

2. Sep 10, 2012

### voko

It is not exactly clear what are all those things you have in the equality, but the second terms looks like a scalar product. Its magnitude is always no greater than the product of the norms of the vectors involved. and one of those seems to be a coordinate unit vector.

3. Sep 10, 2012

### naaa00

Hello Voko,

That's a part of the proof that shows that a map is continous differentiable iff the partials exist and are continous on E. The " D_j f_i " correspond to the partials.

I'm still not sure about the last term. What I'm thinking is that f'(x) and f'(y) are not vectors, but numbers. The thought was that e_j was factored out from both f'(y) and f'(x) - e_j is, indeed, the coordinate unit vector. Then that product isn't a scalar product. Lastly, you got a vector, and claim that the norm of that vector is less than the magnitude of the product (in the second term.) But, then, why they keep the same notation for the f'(x) and f'(y)?

4. Sep 10, 2012

### voko

Seeing all the symbols defined would be helpful. I am not sure what $D_j f_i$ really mean. Partial derivatives would only have one index, but these have two. Are they second-order partial derivatives? Or are they derivatives of components of vector $f(x)$? What is $f'(x)$?

Without this knowledge I could say only very generically that the middle term might be a simple product of a number with a vector, or a scalar product. In either case, if the vector is a unit vector, the inequality follows.

5. Sep 10, 2012

### naaa00

Do you have Rudin's "Principles of mathematical analysis" 3th edition? Rudin's definition of the partials is on pag. 215. And the complete proof of the part avobe is on pag. 219.

The scalar product isn't supposed to be the product of two equal lenght sequences of numbers?

6. Sep 10, 2012

### voko

Okay. $\mathbf{f}(\mathbf{x})$ really is a vector function, and $\mathbf{f}'(\mathbf{x})$ is its differential, that is, a linear transformation (a matrix). The middle term must in fact be $| [\mathbf{f}'(\mathbf{y}) - \mathbf{f}'(\mathbf{x})] \mathbf{e}_j \cdot \mathbf{u}_i|$, and because of the scalar product property I have already mentioned (see Theorem 1.37 on page 16, part (d)), $| [\mathbf{f}'(\mathbf{y}) - \mathbf{f}'(\mathbf{x})] \mathbf{e}_j \cdot \mathbf{u}_i| \le| [\mathbf{f}'(\mathbf{y}) - \mathbf{f}'(\mathbf{x})] \mathbf{e}_j |$, because $\mathbf{u}_i$ is a unit vector, as remarked in the proof; this is the middle term given in the book, and the next inequality follows from definition (c) on page 208.

7. Sep 10, 2012

### naaa00

The LaTex code is not working, apparently... Really difficult to figure out what you wrote.

8. Sep 10, 2012

### voko

Just give it some time; they will fix it. I really don't feel like retyping that again :)