To prove the existing and uniqueness of a solution

Question

Let function $f$ be differentiable and convex in $R^{n}$. How can it be proved that $\forall \lambda > 0$ solution of system equations $f'(x) = -\lambda x$ exists exclusively ($\exists \hspace{3mm} and \hspace{3mm}!$).

Answer 1

It is clear that $f(x)=-\frac{\lambda}{2}|x|^2$ satisfies $(\nabla f)(x)=f'(x)=-\lambda x.$

Suppose that $f$ and $g$ satisfy the system of equations. Then

$$\nabla(f-g)(x)=\nabla f(x)-\nabla f(x)=f'(x)-g'(x)=0,$$ frome where you have that $f-g$ has to be constant. So if $f$ is a solution of the system then $f+c$ is also a solution for any constant $c.$ Thus, there are infinitely many solutions.

However, all the solutions are not convex.

Answer 2

The equation $\nabla f(x) + \lambda x=0$ is the necessary optimality condition of the minization problem $$ \min f(x) + \frac \lambda2 \|x\|^2. $$ Since $f$ is convex, the function $x\mapsto f(x) + \frac \lambda2 \|x\|^2$ is strictly convex for $\lambda>0$. Hence the minimization problem is uniquely solvable, which is equivalent to the unique solvability of $$ \nabla f(x) + \lambda x=0 $$