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By definition, a function $f: U\rightarrow \mathbb R$ defined on an open set $U\subset \mathbb R^n$ is differentiable at $a\in U$ if there is a linear transformation $df_a: \mathbb R^n\rightarrow \mathbb R$ such that

$$\lim_{x\to a} \frac{f(x)-f(a)-df_a(x-a)}{|x-a|}=0.$$

Using this definition I'd like to show that if $f, g: U\rightarrow \mathbb R$ is differentiable at $a\in U$, then $f\cdot g: U\rightarrow \mathbb R$ is also differentiable at $a$ and

$$d(f\cdot g)_a=f(a)\cdot dg_a+df_a\cdot g(a).$$

I thought I should do something like:

\begin{align*} (f\cdot g)(x)&-(f\cdot g)(a)-(f(a)\cdot dg_a(x-a)+df_a(x-a)\cdot g(a))&\\ &=f(x)\cdot g(x)-f(a)\cdot g(a)-f(a)\cdot dg_a(x-a)-df_a(x-a)\cdot g(a)\tag{1}\\ &= {\color{red}{f(x)\cdot g(x)}}-{\color{red}{f(a)\cdot g(x)}}+{\color{green}{f(a)\cdot g(x)}}\\ &-{\color{green}{f(a)\cdot g(a)}}-{\color{green}{f(a)\cdot dg_a(x-a)}}-df_a(x-a)\cdot g(a)\\ &+df_a(x-a)\cdot g(x)-{\color{red}{df_a(x-a)\cdot g(x)}}\\ &={\color{red}{[f(x)-f(a)-df_a(x-a)]\cdot g(x)}}+{\color{green}{f(a)\cdot [g(x)-g(a)-dg_a(x-a)]}}\\ &+df_a(x-a)\cdot (g(x)-g(a)). \end{align*} The problem is the term $df_a(x-a)\cdot (g(x)-g(a))$ which I can't get rid off.

I feel that if I add and subtract the right terms in the line $(1)$ it will be straightforward but I still haven't found those terms.

Any tips?

Thanks.

Remark. I could finish the proof as follows:

$$\frac{df_a(x-a)}{|x-a|}=df_a\left(\frac{x-a}{|x-a|}\right)\leq \max_{|v|=1}df_a(v),$$ implies

$$\lim_{x\to a} \frac{df_a(x-a)}{|x-a|}\cdot (g(x)-g(a))=0$$

since $g(x)\to g(a)$ and $x\mapsto \frac{df_a(x-a)}{|x-a|}$ is bounded. Hence:

Hence:

\begin{align*} &\lim_{x\to a} \frac{(fg)(x)-(fg)(a)-f(a)dg_a(x-a)-df_a(x-a)g(a)}{|x-a|}\\ &=\lim_{x\to a} \frac{f(x)-f(a)-df_a(x-a)}{|x-a|}\lim_{x\to a} g(x)+f(a)\lim_{x\to a} \frac{g(x)-g(a)-dg_a(x-a)}{|x-a|}+\lim_{x\to a}\frac{df_a(x-a)}{|x-a|}\cdot [g(x)-g(a)]=0. \end{align*}

Bernard
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PtF
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  • Note that $\frac{|df_a(x-a)\cdot (g(x)-g(a))|}{|x-a|}\leq |df_a|\cdot |g(x)-g(a)|$ (here $|df_a|$ denotes the operator norm). Since $g$ is differentiable at $a$, it is continuous at $a$, so the second term vanishes as $\to a$. – peek-a-boo Nov 20 '21 at 16:30
  • I was just writing that, but I'd like to avoid that argument. – PtF Nov 20 '21 at 16:32
  • You want to avoid a direct argument which proves what you're asking for? Then what kind of reason are you looking for? – peek-a-boo Nov 20 '21 at 16:38
  • I can explain that.For instance, when you prove that the limit of a product of two functions is the product of the limits you can do that using the fact that if the limit exists at a point then the function is bounded in a neighbourhood of that point.But you can avoid that using a trick, as I did https://math.stackexchange.com/questions/4171310/proof-writing-varepsilon-delta-proof-of-the-product-rule-for-limits-a/4171349#4171349. Something analogous happens when you deal with sequences.So I believe I can avoid using the boundedness argument also in this case. Maybe I can't, but I have to try. – PtF Nov 20 '21 at 16:47
  • Well, you didn't really avoid boundedness. By saying $|f(x)-L|<\min(1,\frac{\epsilon}{3(1+|L|)})$ you're already invoking boundedness (because you're imposing that it is less than $1$). And anyway, even if you don't use the minimum of $1$ and that normalized $\epsilon$, you could always without loss of generality suppose $0<\epsilon<1$, and then multiply then you'd end up with an $\epsilon^2$ term which is less than $\epsilon$. But even in this argument, we're invoking some sort of control of bounds (that squaring small numbers yields a smaller number). So again, I'm not sure what you're after – peek-a-boo Nov 20 '21 at 16:50
  • As I see, the inequality $|f(x)-L|<\min\left{1, \frac{\varepsilon}{3(1+|L|)}\right}$ is not a consequence of boundedness, you're just invoking the definition of $\lim_{x\to a} f(x)=L$ which is prior to boundedness (as you can take any $\varepsilon>0$ in $|f(x)-L|<\varepsilon$). Further, you prove the boundedness around the point $a$ using the definition of limit. What I'm looking for is to avoid arguments which go beyond the definition. Above, when I write $\frac{df_a(x-a)}{|x-a|}\leq \max_{|v|=1}|df_a(v)$ I'm using a lot things beyond my original definition. – PtF Nov 20 '21 at 17:31
  • Right, what I'm saying is the definition of limits tells you in particular that $|f(x)-L|<1$. This is precisely telling you $f$ is bounded (in a neighborhood), and that is exactly what you're using in the rest of your argument. People usually say "oh since $f$ is continuous at $a$, it is bounded in a neighborhood at $a$, therefore, this term is a bounded function times something going to zero, hence the whole thing vanishes". Whereas in your "new proof", you didn't avoid the typical boundedness argument; you just subsumed it as part of your $\epsilon$-$\delta$ proof. – peek-a-boo Nov 20 '21 at 17:34
  • i.e the place where people invoke boundedness, is exactly what you've written out more explicitly from scratch. So, you're not avoiding it, you're just saying in a more "from-scratch" way. So, it's not really a new/different idea for proving that limit of product equals product of limits. I could have a theorem which I prove using 10 lemmas, while someone else might write the exact same proof without separating it out into 10 lemmas. That doesn't make the proofs different. It's just a slightly different presentation. – peek-a-boo Nov 20 '21 at 17:36
  • Conceptually, you're right, it is not different, but it is more clear to understand when you don't have to mention "other facts", as though those "other facts" are equivalent to what you're saying, as it does not require further thinking. We can prove a lot results using equivalent arguments, but some arguments are clearer than others. – PtF Nov 20 '21 at 17:38
  • When someone reads "since $f$ is continuous at $a$ it is bounded around $a$", probability, he will look for the result which proves it. It is pretty much straightforward to say "by definition of continuity we can take this $\varepsilon$". – PtF Nov 20 '21 at 17:43
  • Sure, but now we're getting into the subjective issue of how to write things up. I can certainly appreciate that "my" way of thinking/writing isn't the same as yours, but this is not a mathematical issue, so I don't think continuing this discussion is going to help things along – peek-a-boo Nov 20 '21 at 17:49

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