Let $f$ be a bounded linear functional on a Hilbert space. Show the function $\|x\|^2+f(x)$ achieves a minimum.

Question

Let $f$ be a bounded linear functional on a Hilbert space. Show the function $\|x\|^2+f(x)$ achieves a minimum, describe the point and the minimum value.

The only way I know of solving something of the sort is to introduce some constant to create a real function so we can take derivatives, but I do not see how to incorporate that trick here.

Answer 1

Presumably this is a real Hilbert space since you are talking about a $\min$.

Let $g(x)= \|x\|^2 + \langle c , x \rangle$.

Note that $g(x) = \|{1 \over 2} c + x\|^2 - \|{1 \over 2} c\|^2$ and hence $g(x) \ge g(-{1 \over 2} c) = - \|{1 \over 2} c\|^2$ for all $x$.