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Good evening to everybody! I have to check if the following two functions are Gâteaux differentiable:

$$ (a)~~f(x_1,x_2)=x_1^2x_2~~~~~~~~~~(b)~~f(x_1,x_2,x_3)=(x_1-x_2)\exp(x_3) $$


(a) Is it enough only to name the Gâteaux derivation $\eta$ in any $x\in\mathbb{R}^2$ for any direction $h\in\mathbb{R}^2$?

My calculation gave me $$ \eta=Df(x)[h]=x_1\cdot (2x_2h_1+x_1h_2). $$

Is that enough or is some additional argumentation necessary here?

(b) I do not come along completely with this. My calculation ended here:

$\lim\limits_{t\to 0}\frac{f(x+th)-f(x)}{t}=\lim\limits_{t\to 0}\left[\frac{x_1\exp(x_3+th_3)}{t}+h_1\exp(x_3+th_3)-\frac{x_2\exp(x_3+th_3)}{t}-h_2\exp(x_3+th_3)-\frac{(x_1-x_2)\exp(x_3)}{t}\right]$

What is the next step here? To my opinion the limits do not exist for the fractions but do exist for the other summands. I am a little bit confused...

Can anybody please help me?

Thank you very much and greetings! math12

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If $f$ is Fréchet differentiable, then $d f (x;h) = Df(x) h$, so your calculation is sufficient. –  copper.hat Jan 28 '13 at 17:07
    
How do I know that (a) is Fréchet differentiable? –  math12 Jan 28 '13 at 17:09
    
Both functions $f$ are smooth. –  copper.hat Jan 28 '13 at 17:22
    
I am sorry, I do not understand what you mean. –  math12 Jan 28 '13 at 17:28
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1 Answer 1

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The Gâteaux differential is just the derivative of the function along the line $t \mapsto x+th$, at the point $t=0$.

In both cases above consider the function $\phi(t) = f(x+th)$, and note that $df(x;h) = \phi'(0)$.

So, for (a), $\phi(t) = (x_1+h_1 t)^2 (x_2+h_2 t)$, from which we obtain $df(x;h) = \phi'(0) = x_1(2 h_1 x_2+h_2 x_1)$, which is what you have above.

For (b), $\phi(t) = (x_1-x_2-h_2 t+h_1 t) e^{(x_3+h_3 t)}$, from which we obtain $df(x;h) = \phi'(0) =(h_1-h_2+h_3(x_1-x_2))e^{x_3}$.

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