# Yet Further Help ... Differentiabilty & Maxima ... Browder, Proposition 8.14 ... ...

#### 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 yet further help in fully understanding the proof of Proposition 8.14 ...

In the above proof by Browder, we read the following:

" ... ... For any $$\displaystyle v \in \mathbb{R}^n$$, and $$\displaystyle t \gt 0$$ sufficiently small, we find (taking $$\displaystyle h = tv$$ above) that $$\displaystyle L(tv) + r(tv) \leq 0$$, or $$\displaystyle Lv \leq r(tv)/t$$, so letting $$\displaystyle t \to 0$$ we have $$\displaystyle Lv \leq 0$$. ... ...

Now ... the above quote implies that

$$\displaystyle \lim_{ t \to 0 } \frac{ r(tv) }{ t } = 0$$ ... ...

But why exactly (formally and rigorously) is this the case ... ... ?

I note that we have that $$\displaystyle \lim_{ h \to 0 } \frac{ r(h) }{ |h| } = \lim_{ t \to 0 } \frac{ r(tv) }{ tv } = 0$$

... but this is (apparently anyway) not exactly the same thing ...

we need to be able to demonstrate rigorously that

$$\displaystyle \lim_{ h \to 0 } \frac{ r(h) }{ |h| } = \lim_{ t \to 0 } \frac{ r(tv) }{ t } = 0$$ ... ...

... but how do we proceed to do this ...?

Hope someone can help ... ...

Peter

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EDIT:

Just noticed that in the above quote, Browder argues that

$$\displaystyle L(tv) + r(tv) \leq 0$$ implies that $$\displaystyle Lv \leq r(tv)/t$$ ... ...

... BUT ... i suspect he should have written

$$\displaystyle L(tv) + r(tv) \leq 0$$ implies that $$\displaystyle Lv \leq - r(tv)/t$$ ... ..

... however ... in either case ... when we let $$\displaystyle t \to 0$$ we get the same result ... namely $$\displaystyle Lv \leq 0$$...

==============================================================================

Last edited:

#### Opalg

##### MHB Oldtimer
Staff member
In the above proof by Browder, we read the following:

" ... ... For any $$\displaystyle v \in \mathbb{R}^n$$, and $$\displaystyle t \gt 0$$ sufficiently small, we find (taking $$\displaystyle h = tv$$ above) that $$\displaystyle L(tv) + r(tv) \leq 0$$, or $$\displaystyle Lv \leq r(tv)/t$$, so letting $$\displaystyle t \to 0$$ we have $$\displaystyle Lv \leq 0$$. ... ...

Now ... the above quote implies that

$$\displaystyle \lim_{ t \to 0 } \frac{ r(tv) }{ t } = 0$$ ... ...

But why exactly (formally and rigorously) is this the case ... ... ?

I note that we have that $$\displaystyle \lim_{ h \to 0 } \frac{ r(h) }{ |h| } = \lim_{ t \to 0 } \frac{ r(tv) }{ tv } = 0$$

... but this is (apparently anyway) not exactly the same thing ...

we need to be able to demonstrate rigorously that

$$\displaystyle \lim_{ h \to 0 } \frac{ r(h) }{ |h| } = \lim_{ t \to 0 } \frac{ r(tv) }{ t } = 0$$ ... ...

... but how do we proceed to do this ...?
This is another case where you have to distinguish between fixed and variable vectors. Here, $v$ is fixed, but $tv$ varies, and goes to $0$ as $t\to0$. So $$\displaystyle \lim_{t\to0}\frac{r(tv)}{t|v|} = 0$$, and then you can multiply by the fixed nonzero scalar $|v|$ to get $$\displaystyle \lim_{t\to0}\frac{r(tv)}{t} = 0$$.

Just noticed that in the above quote, Browder argues that

$$\displaystyle L(tv) + r(tv) \leq 0$$ implies that $$\displaystyle Lv \leq r(tv)/t$$ ... ...

... BUT ... i suspect he should have written

$$\displaystyle L(tv) + r(tv) \leq 0$$ implies that $$\displaystyle Lv \leq - r(tv)/t$$ ... ..

... however ... in either case ... when we let $$\displaystyle t \to 0$$ we get the same result ... namely $$\displaystyle Lv \leq 0$$...
Absolutely correct! Browder has omitted a minus sign. But his conclusion is correct.

#### Peter

##### Well-known member
MHB Site Helper
This is another case where you have to distinguish between fixed and variable vectors. Here, $v$ is fixed, but $tv$ varies, and goes to $0$ as $t\to0$. So $$\displaystyle \lim_{t\to0}\frac{r(tv)}{t|v|} = 0$$, and then you can multiply by the fixed nonzero scalar $|v|$ to get $$\displaystyle \lim_{t\to0}\frac{r(tv)}{t} = 0$$.

Absolutely correct! Browder has omitted a minus sign. But his conclusion is correct.

HI Opalg ...