I'm trying to prove that if $f,g$ are two functionals on a Banach space $X$ that have the same Gateaux derivative for all $x \in X$, then $f-g = \text{constant}$. I can show that under the hypotheses, the Gateaux derivative of $f-g$ must be $0$ for all $x \in X$, but I don't know how to prove that this implies $f-g$ is a constant. I can't seem to generalize the proof from ordinary calculus.
Asked
Active
Viewed 115 times
1
-
Do you know a MVT for the Gateaux differentiation ? – Tony Piccolo Jul 23 '14 at 23:35
-
My text only proves an MVT for the Frechet derivative, but I suppose I should be able to prove one for Gateaux as well. – QuantumDots Jul 24 '14 at 13:21
1 Answers
0
Suppose that the functional $F$ has a Gâteaux derivative at every point of a Banach space.
If $\varphi(t)=F(x+th)$ then $$\varphi'(t)=\lim_{s \rightarrow 0} \frac {\varphi (t+s)-\varphi(t)}s=\lim_{s \rightarrow 0} \frac {F(x+th+sh)-F(x+th)}s=DF(x+th)h$$ (pay attention: $DF(x+th)h\,$ is not a product)
By the MVT for real functions of one variable applied to $\varphi$, one has $$F(x+h)-F(x)=\varphi(1)-\varphi(0)=\varphi'(\xi)=DF(x+\xi h)h$$ for some $\xi \in \,]0,1[\,$.
Use this result.
Tony Piccolo
- 4,426