Functional in Hilbert space

Question

Let $H$ be a Hilbert space and $0\neq x\in H$. I want to prove that there is an unique $f\in H^*$, such that $\|f\|=1$ and $f(x)=\|x\|$.

Any ideas on how to approach this problem

Answer 1

Let $K=\Bbb R\text{ or } \Bbb C$ the field of scalars. Let $V$ be the subspace of $H$ defined by $$V=\{\lambda x\}_{\lambda\in K}$$ Consider the continuous form $\phi\in V^*$ defined by $$\phi(\lambda x)=\lambda\space ||x||$$

It is clear that $\phi( x)= ||x||$.

Besides, for all $\lambda x\in V$ one has $$|\phi(\lambda x)|=|\lambda|\space ||x||=||\lambda x||=||x||$$ Hence $$||\phi||_V=1$$Now by Hahn-Banach theorem we can extend $\phi$ to the whole $H$ with preservation of the norm.

(We get this way a form $f\in H^*$ such that the restriction $f_{|V}=\phi$ and $||f||=||\phi||=1$)

Answer 2

Call such functional $f$ norming. Let us try to convince us that Riesz's theorem is not quite necessary here.

Let $X$ be a Banach space and $x\in X$ be a non-zero vector. Then $x$ has a unique norming functional whenever the function $t\mapsto \|x+ty\|$ is differentiable at 0 for every $y\in X$. Please see here for more hints. In our case (that is for Hilbert spaces), this function is certainly differentiable at 0 given the shape of the unit sphere of the two-dimensional Euclidean space.

Here $f$ can be found explicitly—take $f=\frac{1}{\|x\|}x$.

Answer 3

[I]. A real or complex Hilbert space $H$ is its own dual. If $f\in H^*$ then there exists $y_f\in H$ such that the inner product <$x,y_f$> is equal to $f(x)$ for all $x\in H.$ And $y_f$ is unique, because if $f(x)=<x, y'_f>$ for all $x,$ then $$\|y_f-y'_f\|^2=<y_f-y'_f,y_f>-<y_f-y'_f,y'_f>=f(y_f-y'_f)-f(y_f-y'_f)=0.$$

Also $\|f\|=\|y_f\|$ because, in the non-trivial case $f\ne 0,$ we must have $y_f\ne 0,$ so $$(i).\quad f(y_f)/\|y_f\|=\|y_f\| \implies \|f\|\geq \|y_f\|.$$

$$(ii).\quad \forall x\in H\;(|f(x)|=|<x,y_f>|\leq \|x\|\cdot \|y_f\|) \implies \|f\|\leq \|y_f\|.$$

[II]. For $0\ne x\in H,$ let $f(y)=$<$y,x/\|x\|$> for all $y\in H.$ Then $f(x)=\|x\|$ and $\|f\|=1.$

Suppose $g\in H^*$ with $g(x)=\|x\|$ and $\|g\|=1.$ Let $g(z)=$<$z,y_g$> for all $z\in H.$ Then $\|y_g\|=\|g\|=1.$ We have $$\|x\|=g(x)=<x,y_g>=|<x,y_g>|\leq \|x\|\cdot \|y_g\|=\|x\|.$$

But the Cauchy-Schwarz Inequality $|$<$u,v$>$|\leq \|u\|\cdot \|v\|$ is a strict inequality unless $u,v$ are linearly dependent. So, since $|$<$x,y_g$>$|=\|x\|=\|x\| \cdot \|y_g\|$ and $x\ne 0\ne y_g,$ there exists scalar r such that $y_g=rx.$ This implies $$0\ne \|x\|=g(x)=<x, rx>$$ so $r=1/\|x\|.$ Hence $y_g=x/\|x\|$ and $g=f.$

[III]. Appendix: To show that $H$ is its own dual: Let $E$ be an orthonormal Hilbert-space basis for $H$. Let $0\ne f\in H^*.$ Then $f$ is uniquely determined by $\{(e,f(e):e\in E\}.$ Let $E_f=\{e\in E: f(e)\ne 0\}.$

Let $F_f$ be the set of finite non-empty subsets of $E_f.$ For $S\in F_f$ let $$x_S=\sum_{e\in S}e\;\overline {f(e)}$$ (where $\overline {z}$ denotes the complex conjugate of $z$). Then $f(x_S)=\sum_{e\in S}|f(e)|^2=\|x_S\|^2 $. Therefore $$\infty >\|f\|\geq \sup_{S\in F_f}\sum_{e\in S}|f(e)|^2=\sum_{e\in E_f}|f(e)|^2.$$ Therefore $y=\sum_{e\in E_f}e\; \overline {f(e)}\in H.$ The functional $g(x)=$<$x,y$> agrees with $f$ for every $x\in E,$ so $g=f.$

Answer 4

My other answer includes extra useful info on Hilbert spaces. This is the short answer.

For Hilbert space $H$ and $0\ne x\in H,$ then $H$ has an orthonormal Hilbert-space basis $E$ with $x/\|x\|\in E.$ For brevity, let $e_0=x/\|x\|.$

For $y=\sum_{e\in E}y_ee\in H$ let $f(y)=y_{e_0}.$ Then $f\in H^*$ with $\|f\|\leq 1,$ as $$|f(y)|^2=|y_{e_0}|^2\leq \sum_{e\in E}|y_e|^2=\|y\|^2.$$ And $f(x)=\|x\|\ne 0,$ so $\|f\|\geq 1.$ Hence $\|f\|=1.$

Suppose $g\in H^*$ with $g(x)=\|x\|$ and $g\ne f.$ We have $g(e_0)=1.$ Now take $e\in E$ with $e\ne e_0$ and $g(e)\ne 0.$ Let $\overline {g(e)}$ denote the complex conjugate of $g(e).$ Then $$\|e_0+e\;\overline {g(e)}\|=\sqrt {1+|g(e)|^2}< 1+|g(e)|^2=|g(\;e_0+e\;\overline {g(e)}\;)|$$ so $\|g\|>1.$