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Can someone confirm if this exercise is correctly done? The context of the exercise is multivariable calculus, that is, $E$ and $F$ are Banach spaces of unknown dimension.

What I did: I must show that if $f\in C^1(E,F)$ and $f(tx)=tf(x)$ for all $t>0$ for all $x\in E\setminus\{0\}$ then $f\in\mathcal L(E,F)$. Using the chain rule I know that

$$\partial[f(tx)]=t\partial f(tx)=t\partial f(x)=\partial[t f(x)]\implies \partial f(tx)=\partial f(x),\quad \forall t>0$$

Thus, by the continuity of $\partial f$, $\partial f(x)$ is a constant function for each $x\in E$, and by differentiation it can be seen that

$$f(x)=\partial f(x)x+v\quad\text{and}\quad f(tx)=tf(x)\implies v=0$$

Thus $f$ is linear, as required.


UPDATE: I now think that the above is not totally correct. I solved the exercise in a different way (with a more explicit result) using the fact that $f$ is differentiable at zero.

From the first attempt above we knows that $\partial f$ is a constant function in the sets $\{tx:t>0\text{ and } x\in E\setminus\{0\}\}$. And because $f$ is differentiable at zero by assumption then using the directional derivatives of $f$ at zero we can see that

$$D_vf(0):=\lim_{t\to 0}\frac{f(0+tv)-f(0)}t\in F\implies\matrix{f(0)=0\;\text{ and }\\D_vf(0)=f(v)=-f(v),\forall v\in E\setminus\{0\}}$$

Thus $f$ is a constant function and because $f(0)=0$ then $f=0$.

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  • $\begingroup$ $f(tx)=tf(x)$ holds only for a particular $t>0$ ? $\endgroup$ – Gabriel Romon Sep 5 '17 at 11:43
  • $\begingroup$ @LeGrandDODOM I edited... it is for all $t>0$. It is positive homogeneity of degree $1$. $\endgroup$ – Masacroso Sep 5 '17 at 11:45
  • $\begingroup$ One more thing, by $f\in\mathcal L(E,F)$, you don't require that $f(x+y)=f(x)+f(y)$ ? $\endgroup$ – Gabriel Romon Sep 5 '17 at 11:59
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Let $x\in E$ be fixed. Define $\gamma: \mathbb R\to E, t\mapsto tx$ and $g:\mathbb R \to F, t\mapsto tf(x)$.

Using the chain rule, note that for all $t,h\in \mathbb R$, $d(f\circ \gamma)(t)(h) = hdf(tx)(x)$.

Let $t>0$ be fixed. Since $f\circ \gamma$ and $g$ coincide on a neighborhood of $t$, their Fréchet derivatives at $t$ are the same: for all $h\in \mathbb R$, $hdf(tx)(x) = hf(x)$. Setting $h=1$ yields $$df(tx)(x) = f(x)$$

The idea is then to let $t$ go to $0$. If all goes well, we end up with $df(0)(x) = f(x)$, that is to say $df(0)=f$, hence $f\in\mathcal L(E,F)$.

Note that $$\begin{align}\|df(tx)(x)-df(0)(x)\|&=\|(df(tx)-df(0))(x)\|\\ &\leq \|df(tx)-df(0)\|\|x\| \\ &\leq \|df\|\|tx\|\|x\|\\ &=t\|df\|\|x\|^2 \end{align}$$

Letting $t\to0$ from above, we get $\lim_{t\to 0+}df(tx)(x) = df(0)(x)$, hence $f(x)=df(0)(x)$.

This holds for all $x\neq 0$, and we're done.

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  • $\begingroup$ @Masacroso see my edits. $\endgroup$ – Gabriel Romon Sep 5 '17 at 12:37
  • $\begingroup$ I see... but still I dont follow what mean the expression $d(f\circ\gamma)(t)(h)$, I understand up to $t$ but, for what function is $h$ a variable? $\endgroup$ – Masacroso Sep 5 '17 at 13:17
  • $\begingroup$ @Masacroso ah sorry if I've been unclear with my notation. $d(f\circ\gamma)(t)$ is the Fréchet derivative of $f\circ\gamma$ at $t$. Remember that the Fréchet derivative at $t$ is a bounded linear operator (so it's a function). $d(f\circ\gamma)(t)$ is therefore a function, and you can apply it at any $h$, yielding $d(f\circ\gamma)(t)(h)$. $\endgroup$ – Gabriel Romon Sep 5 '17 at 13:29
  • $\begingroup$ but $d(f\circ\gamma)(t)$ is a linear function with domain $E$, but $h$ is a real number in your definition. $\endgroup$ – Masacroso Sep 5 '17 at 13:42
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    $\begingroup$ @Masacroso No, $f\circ \gamma :\mathbb R \to F$, hence $d(f\circ\gamma)(t): \mathbb R \to F$. $\endgroup$ – Gabriel Romon Sep 5 '17 at 13:44

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