Frechet Derivative simple application

Question

Suppose that $X$ is a square integrable random variable, i.e. $X\in L^2(\Omega,\mathcal{F},\mathbb{P})$ and the function $g:L^2(\Omega,\mathcal{F},\mathbb{P})\rightarrow\mathbb{R}$ where $$g(X(\omega))=-\alpha X(\omega)^2+\beta X(\omega)$$ where $\alpha$, $\beta$ are positive constants. Could someone apply the Frechet derivative to this function and if yes how what is the method to do it? I am trying to understand what is the difference between Frechet derivative and the classic one that we use from our school years and how can we apply it. I only know that in Banach spaces we can use the Frechet derivative and since $L^2$ is a Banach space I want to know how can this derivative de implemented in my case

Answer 1

I think you're trying to define the function $g:L^2(\Omega,\mathcal{F},\mathbb{P})\to L^1(\Omega,\mathcal{F},\mathbb{P})$ such that $g(X)=-\alpha X^2+\beta X$. Since $X$ is a random variable, I assume that $(\Omega,\mathcal{F},\mathbb{P})$ is a probability space, then we have $L^2\subseteq L^1$. Let's see that $g$ is well defined, that is, $g(X)\in L^1$ when $X\in L^2$. Indeed $$\int_\Omega|g(X)|d\mathbb{P}\leq\alpha\int_\Omega|X|^2d\mathbb{P}+\beta\int_\Omega|X|d\mathbb{P}<\infty.$$ The integrals are finite because $X\in L^2\subseteq L^1$.

Let's compute the Fréchet derivative. Let $X,H\in L^2$ be random variables, then \begin{align}g(X+H)&=-\alpha(X+H)^2+\beta(X+H)\\ &=-\alpha X^2+\beta X-2\alpha HX+\beta H-\alpha H^2\\ &=g(X)+(\beta-2\alpha X)H-\alpha H^2.\end{align}

Notice that the map $L:L^2\to L^1$ given by $L(H)=(\beta-2\alpha X)H$ is clearly linear. Let's prove that is continuous: \begin{align} \|(\beta-2\alpha X)H\|_{L^1}&\leq\int_\Omega|\beta-2\alpha X|\cdot|H|d\mathbb{P}\\ &\leq\beta\int_\Omega|H|d\mathbb{P}+2\alpha\int_\Omega|XH|d\mathbb{P}\\ &\leq\beta\|H\|_{L^1}+2\alpha\|X\|_{L^1}\|H\|_{L^1}\\ &\leq\left(\beta+2\alpha\|X\|_{L^1}\right)\|H\|_{L^2} \end{align} where the third inequality is because of Holder inequality and the last inequality is because of Cauchy-Schwarz $$\|H\|_{L^1}=\int_\Omega|H|d\mathbb{P}\leq\left(\int_\Omega d\mathbb{P}\right)^{1/2}\left(\int_\Omega H^2d\mathbb{P}\right)^{1/2}=\|H\|_{L^2}.$$

Lastly, let's check that $-\alpha H^2\in o(\|H\|_{L^2})$. Since $\|-\alpha H^2\|_{L^1}\leq\alpha\int_\Omega H^2d\mathbb{P}\leq\alpha\|H\|_{L^2}^2$, then $$\frac{\|-\alpha H^2\|_{L^1}}{\|H\|_{L^2}}\leq\|H\|_{L^2}\xrightarrow{\|H\|_{L^2}\to 0}{0}.$$

With all this we have $g(X+H)=g(X)+L(H)+o(\|H\|_{L^2})$ and conclude that $g$ is Fréchet differentiable in $X$ and $$Dg(X)=(\beta-2\alpha X)\in\mathcal{L}(L^2,L^1).$$