# Differentiabilty and Continuity of Vector-Valued Functions ... Browder, Proposition 8.13 ... ...

#### Peter

##### Well-known member
MHB Site Helper
I am reading Andrew Browder's book: "Mathematical Analysis: An Introduction" ... ...

I am currently reading Chapter 8: Differentiable Maps and am specifically focused on Section 8.2 Differentials ... ...

I need some further help in fully understanding the proof of Proposition 8.13 ...

I think that a fully detailed proof of Proposition 8.13 reads somewhat as follows:

Browder's Definition 8.9 essentially means that $$\displaystyle \text{df}_p$$ exists if

$$\displaystyle \lim_{ h \to 0 } \frac{1}{| h | } (f(p + h) - f(p) - \text{df}_p h ) = 0$$

Thus ... if we take $$\displaystyle \epsilon = C - \| df_p \|$$ then we can find $$\displaystyle \delta$$ such that ...

$$\displaystyle | | h | \lt \delta \Longrightarrow \frac{1}{ | h | } | (f(p + h) - f(p) - \text{df}_p h ) - 0 | \leq \epsilon$$ ... ... ... (1)

so that

$$\displaystyle | h | \lt \delta \Longrightarrow | (f(p + h) - f(p) - \text{df}_p h ) | \leq \epsilon | h |$$ ... ... ... (2)

Now the reverse triangle inequality (Duistermaat & Kolk Lemma 1.1.7 (iv) ) states that

$$\displaystyle \| x - y \| \geq | \ \| x \| - \| y \| \ |$$

Using the reverse triangle inequality we have

$$\displaystyle | (f(p + h) - f(p) ) - ( \text{df}_p h ) | \geq | \ | f(p + h) - f(p) | - | \text{df}_p h ) | \ |$$ ... ... ... (3)

Now (2) and (3) $$\displaystyle \Longrightarrow$$

$$\displaystyle | \ | f(p + h) - f(p) | - | \text{df}_p h ) | \ | \leq \epsilon |h|$$

$$\displaystyle \Longrightarrow | f(p + h) - f(p) | \leq | \text{df}_p h | + \epsilon |h|$$

Now $$\displaystyle | \text{df}_p h ) | \leq | \text{df}_p | | h ) |$$ (Is that correct? )

... so that ...

$$\displaystyle | f(p + h) - f(p) | \leq | \text{df}_p | | h ) | + \epsilon |h|$$

$$\displaystyle \Longrightarrow | f(p + h) - f(p) | \leq ( \| \text{df}_p \| + \epsilon ) |h| + C |h|$$

Is that correct?

Now ... my specific problem is how to rigorously and validly make the move

$$\displaystyle | \ | f(p + h) - f(p) | - | \text{df}_p h ) | \ | \leq \epsilon |h|$$

$$\displaystyle \Longrightarrow | f(p + h) - f(p) | \leq | \text{df}_p h | + \epsilon |h|$$

... ... since I have effectively ignored the modulus signs around $$\displaystyle \ | f(p + h) - f(p) | - | \text{df}_p h ) |$$ ...

... that is I have assumed that $$\displaystyle | f(p + h) - f(p) | \geq | \text{df}_p h ) |$$ ....

Can someone please explain how i deal with this issue ...

Help will be much appreciated ...

Peter

Last edited:

#### Opalg

##### MHB Oldtimer
Staff member
Now ... my specific problem is how to rigorously and validly make the move

$$\displaystyle | \ | f(p + h) - f(p) | - | \text{df}_p h ) | \ | \leq \epsilon |h|$$

$$\displaystyle \Longrightarrow | f(p + h) - f(p) | \leq | \text{df}_p h | + \epsilon |h|$$

... ... since I have effectively ignored the modulus signs around $$\displaystyle \ | f(p + h) - f(p) | - | \text{df}_p h ) |$$ ...

... that is I have assumed that $$\displaystyle | f(p + h) - f(p) | \geq | \text{df}_p h ) |$$ ....

Can someone please explain how i deal with this issue ...
For real numbers $X$ and $Y$, $|X|\leqslant Y$ means $-Y\leqslant X\leqslant Y$. In particular, $|X|\leqslant Y \Longrightarrow X\leqslant Y$.

In this case, $| f(p + h) - f(p) | - | \text{df}_p h | \leqslant\bigl| \ | f(p + h) - f(p) | - | \text{df}_p h | \bigr| \leqslant | \text{df}_p h ) |$.

#### Peter

##### Well-known member
MHB Site Helper
For real numbers $X$ and $Y$, $|X|\leqslant Y$ means $-Y\leqslant X\leqslant Y$. In particular, $|X|\leqslant Y \Longrightarrow X\leqslant Y$.

In this case, $| f(p + h) - f(p) | - | \text{df}_p h | \leqslant\bigl| \ | f(p + h) - f(p) | - | \text{df}_p h | \bigr| \leqslant | \text{df}_p h ) |$.

Thanks for a most helpful Post, Opalg

Peter