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Suppose that $f,g: U\to \mathbb{K}$ where $U$ is an open set around $a$ are differentiable in $a$. Show that $<f,g>: U \to \mathbb{K}$ where $<f,g>(x) = <f(x),g(x)>$ is also differentiable i $a$.

I know $f$ is differentiable in $a$ iff there is a $T \in L(\mathbb{K},E)$ where $E$ is a vector space such that \begin{align} \lim_{x \to a} \frac{f(x)-f(a)-T(x-a)}{\Vert x-a \Vert} = 0 \end{align}

Here I claim that $<f,g>=<f',g>+<f,g'>$. I think I have to show that $<f',g>+<f,g'>$ is linear, right? But I don't know how to show that either.Because for $\lambda\in \mathbb{K}$ $$<f',g>+<f,g'>(\lambda x)=<f'(\lambda x),g(\lambda x)> + <f(\lambda x),g'(\lambda x)>=\lambda<f'(x),g(\lambda x)>+\lambda<f(\lambda x),g'(x)>$$

But I don't know how to do the rest. Would someone please help? I have an exam tomorrow and I'm in a kind of hurry.

daw
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Mina
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    Very similar to the proof of product rule for differentiation. – Kavi Rama Murthy Nov 09 '21 at 11:53
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    If you know $T_f, T_g$ such that the limits for $f, g$ exist and are equal to $0$, can you construct a $T_{\langle f, g \rangle}$ which will guarantee that the same holds for $\langle f, g \rangle$? – πr8 Nov 09 '21 at 11:54
  • @πr8 I'm getting so confused. I don't really know how to show that. Also I think I have to show that $<f',g>+<f,g'>$ is linear? And I don't know how to show that either. – Mina Nov 09 '21 at 11:59
  • @KaviRamaMurthy In fact, that is the product rule. – Michael Hoppe Nov 09 '21 at 12:20
  • @MichaelHoppe Yes, I think it's similar. But how can I show this with this limit? – Mina Nov 09 '21 at 12:31
  • Do it just like Leibniz did in former times: add and subtract $\langle f(a),g(x)\rangle>$ in the numerator. – Michael Hoppe Nov 09 '21 at 12:41

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By assumption, you can write that

\begin{align} f(x) &= f(a) + T_f(x-a) + \varepsilon_f (x; a) \\ g(x) &= g(a) + T_g(x-a) + \varepsilon_g (x; a) \end{align}

where $\varepsilon_f, \varepsilon_g$ vanish at $a$ (i.e. as $x \to a$) faster than $\| x - a \|$. Now, write

\begin{align} \langle f(x), g(x) \rangle &= \langle f(a) + T_f(x-a) + \varepsilon_f (x; a), g(a) + T_g(x-a) + \varepsilon_g (x; a) \rangle \\ &= \langle f(a), g(a) \rangle + \langle T_f(x-a), g(a) \rangle + \langle f(a), T_g(x-a) \rangle + E_{f,g} (x; a) \end{align}

where $E_{f,g} (x; a)$ can be written out explicitly, and vanishes at $a$ faster than $\| x - a \|$ (this is worth checking).

So, for $h = \langle f, g \rangle$, you can write that \begin{align} h(x) &= h(a) + \langle T_f(x-a), g(a) \rangle + \langle f(a), T_g(x-a) \rangle + E_{f,g} (x; a) \\ &= h(a) + \langle T_f^*g(a) + T_g^*f(a), x-a \rangle + E_{f,g} (x; a). \end{align}

It is thus natural to write $T_h = \left( T_f^*g(a) + T_g^*f(a) \right)^*, \varepsilon_h(x;a) = E_{f,g} (x; a)$ to obtain a similar expression to before:

\begin{align} h(x) &= h(a) + T_h(x-a) + \varepsilon_h (x; a) \end{align}

where $\varepsilon_h$ vanishes at $a$ faster than $\| x - a \|$, which is precisely what is needed to prove the differentiability of $h$ at $a$.

πr8
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