An exercise about variational principle Let $H$ be a Hilbert space. Let $l: H \to \mathbb{R}$ be a continuously linear function. Let $g: H \to \mathbb{R}$ be defined by $$g \left ( x \right )= \frac{\left \| x \right \|^2}{2}-l\left ( x \right )$$ for all $x$ in $H$. Prove that there is $u \in H$ such that for all $x \in H$ we get $l \left ( x \right ) = \langle x, u \rangle$, where $\langle .,. \rangle$ is the scalar product in $H$.

My friend has a solution for it: Since $l$ is linear continuous, then $l$ is bounded. It follows that $g$ is coercive, i.e, $g\left ( x \right ) \to \infty$ as $\left \| x \right \| \to \infty $ and therefore attains a (global) minimum at , say $u$. Then $$0=g'\left ( u \right ) = u-l'\left ( u \right )$$ And therefore for all $x \in H$ we have $l\left (x \right ) = \langle x,u \rangle$.

I'm still in dark about some problems:


*

*I don't understand the derivative of scalar product function, for example $l\left (x \right )$ as above.

*Why we get that if we have $l'\left ( u \right )=u$ then $l\left ( x \right )=\langle x,u \rangle$ ?



Help me please. Thank you so much for help.
 A: It's easier to understand if you remember that $ dg $ is a linear functional:
$$
dg_x(v) = \langle x, v \rangle - l(v) 
$$
here $ v \in T_xH \simeq H $ is a tangent vector at $ x $. On the right side, use the fact that the derivative of a bounded linear operator is itself.
In other words:
$$
g(x+v) = dg_x(v) + O(|v|^2)
$$
Now the statement that $ g $ has a minimum at $ u $ implies $ dg_u = \langle u, \cdot \rangle - l(\cdot)  = 0$.
EDIT: By the way, this is a special case of the Riesz Representation theorem.
A: An answer that is equivalent to your friend's answer, but without technically invoking the concept of differentiation on normed linear spaces:
For $x\in H$, consider the function $h_x:\mathbb{R}\rightarrow\mathbb{R}$ defined by $h_x(t) = g(u+tx)$. Then $h_x$ has a minimum at $t=0$, so if $h_x$ is differentiable there, then $h_x'(0)=0$, i.e. if the limit $\lim\limits_{t\rightarrow 0}{\frac{g(u+tx)-g(u)}{t}}$ exists, then it is equal to zero.
Now
\begin{align*}g(u+tx)-g(u) &= \frac{\|u+tx\|^2}{2}-l(u+tx) - \frac{\|u\|^2}{2}+l(u) \\
&= \frac{1}{2}(\langle u+tx,u+tx\rangle-\langle u,u\rangle)-(l(u+tx)-l(u)) \\
&=t\langle u,x\rangle + \frac{t^2}{2}\langle x,x\rangle - tl(x)
\end{align*}
so
$$ \lim\limits_{t\rightarrow 0}{\frac{g(u+tx)-g(u)}{t}} = \lim\limits_{t\rightarrow 0}{\left(\langle u,x\rangle + \frac{t}{2}\langle x,x\rangle - l(x)\right)} = \langle u,x\rangle - l(x). $$
Hence, $h_x$ is differentiable at $0$, so the above limit is equal to zero, yielding your result.

Remark: For $f:H\rightarrow\mathbb{R}$, we can define a notion of differentiability as follows: We say that $f$ is differentiable at $x\in H$ if there exists a bounded linear operator $Df_x: H\rightarrow\mathbb{R}$ such that
$$ \lim\limits_{h\rightarrow 0}{\frac{\|f(x+h)-f(x)-Df_x(h)\|}{\|h\|}} = 0 $$
(the limit is taken for $h\in H$). This notion of differentiability carries over many of the notions of differentiability on Euclidean space, and in particular the following two properties will be useful for this problem:


*

*If $f$ is differentiable at $x$, and it also attains a local extremum at $x$, then $Df_x = 0$.

*If $f$ is linear, then for every $x\in H$ we have $Df_x = f$.


One can check that the derivative of $x\mapsto\frac{\|x\|^2}{2}$ at $x$ is $h\mapsto\langle h,x\rangle$, which yields your result.
See Fréchet derivative for more details.
