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 $f=0$.

  • $\begingroup$ $f(tx)=tf(x)$ 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

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.

  • $\begingroup$ @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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.