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Let $f:[a,b]\to E$ where $E$ is a Banach space and let $Df(x,h)$ be its Gâteaux differential in $x$ with direction $h$. If $\mathbb{R}\to E$, $h\mapsto Df(x,h)$ is linear and continuous, then we write $f_c'(x)h:=Df(x,h)$.

I suppose that I have been able to prove the two following facts to myself, but my textbook, Kolmogorov-Fomin's Элементы теории функций и функционального анализа says nothing about it explicitly, and I find nothing on line:

  • If $f$ is continuous on $[a,b]$ and $F(x):=\int_a^x f(t)dt$ is the integral function defined by the Riemann integral, then $DF(x,h)=hf(x)$. To prove this, I have essentially used the same tools used for the fundamental theorem of calculus.
  • If $f_c':[a,b]\to \mathscr{L}(\mathbb{R},E)$ is continuous on $[a,b]$ then $h(f(b)-f(a))=\int_a^b f_c'(t)hdt$ (since $h$ is a scalar, we can consider $f_c'$ as a vector of $E$, but I like better to consider it as a map, as it happens in general). To prove this, I have used the finite increment formula (p. 483 (10) here).

Are these results correct? I thank you very much for any answer!

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The second statement is essentially proven few pages later, p. 487 in your link, since in the case of functions defined on the real line Gâteaux derivative coincides with Fréchet. The first one follows from the estimate (which I assume is the same you used): $$ \left\|\frac{F(x+th)-F(x)}{t}-hf(x)\right\|=\frac{\left\|\int\limits_{x}^{x+th}(f(s)-f(x))ds\right\|}{t}\leqslant |h|\max\limits_{x\le s\le x+th}\|f(s)-f(x)\|\to 0, $$ as $t\to 0.$

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  • $\begingroup$ I definitely agree with Rényi in thinking that "If I feel unhappy, I do mathematics to become happy. If I am happy, I do mathematics to keep happy", but, if I see that my mathematical thought aren't incorrect, I become even happier. I ∞-ly thank you ...and welcome to MSE! $\endgroup$ – Self-teaching worker Dec 17 '14 at 20:49

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