Quadratic Functional Differentiability

Question

I would like to solve the following:

Let $T$ be a self-adjoint bounded operator on a Hilbert space $H$. Consider the quadratic functional $\Phi$ defined by:

\begin{equation} \Phi(x)=\frac{1}{2}(Tx,x) \end{equation}

1. Show that $\Phi$ is differentiable and $\nabla \Phi=T$.
2. Show that $\Phi$ is convex if $T$ is strictly positive.

Answer 1

$$ \begin{align} 2\Phi(x+h)-2\Phi(x) & = (T(x+h),x+h)-(Tx,x) \\ & = (Tx,x)+(Tx,h)+(Th,x)+(Th,h)-(Tx,x) \\ & = (Tx,h)+(Th,x)+(Th,h) \\ & = (Tx,h)+(h,Tx)+(Th,h) \end{align} $$ I assume you must be working on a real space, and working with an operator $T$ defined on the whole space, which makes it bounded because it is symmetric. If so, then the linear approximation to $\Phi$ is $$ \Phi(x+h)=\Phi(x)+(Tx,h)+o(\|h\|) $$