What is $\| f \|$ where $f(x)=\sum\limits_{n=1}^\infty \frac{1}{3^n} \langle x,e_n\rangle$

Question

Question:

1) Let $\{ e_n : n\in \mathbb{N}\}$ be an ortonormal basis of a Hilbert space $H$.If $$x\mapsto f(x)=\sum_{n=1}^\infty \frac{1}{3^n} \langle x,e_n\rangle$$ for all $x \in H$ , then determine $\| f \| \text{?}$

2) Every orthonormal set in a Hilbert space $H$ must be closed in $H$. (T\F)

work: 1) I find $f(e_n)=1/3^n,$ after that I cant manage. Please help.

2) I think 2 is false, as orthonormal set are dense in $H.$ Please help

Answer 1

1) Look up @JimmyK4542 for the bound.

Now put $x=\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{3^{n}}e_{n}$, this is defined since $\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{3^{n}}\|e_{n}\|=\sum_{n=1}^{\infty}\dfrac{1}{3^{n}}=\dfrac{1}{2}<\infty$ and this attains the bound.

2) If this question were asking that if the span of the orthonormal set is closed, then look up any Hilbert space with Hamel basis of cardinality $\mathfrak{c}$.

If the question were merely asking that if the orthonormal set is closed, then yes, as indicated by @Kavi Rama Murthy.

Answer 2

user284331 has already given a hint for problem 2.

For 1, note that we can use the Cauchy-Schwarz inequality to get $$|f(x)| = \left|\sum_{n = 1}^{\infty}\dfrac{1}{3^n}\langle x,e_n \rangle\right| \le \left(\sum_{n = 1}^{\infty}\dfrac{1}{3^{2n}}\right)^{1/2} \cdot \left(\sum_{n = 1}^{\infty}\left|\langle x,e_n \rangle\right|^2\right)^{1/2} = \dfrac{1}{2\sqrt{2}}\|x\|$$ for all $x \in H$.

Equality holds if there exists a constant $C$ such that $\langle x,e_n \rangle = \dfrac{C}{3^n}$ for all $n$, i.e. $x = \displaystyle\sum_{n = 1}^{\infty} \ \dfrac{C}{3^n} \cdot e_n$.

From this, we get that $\|f\| = \dfrac{1}{2\sqrt{2}}$.

Answer 3

You are right for the second question . you can in fact consider any finite orthonormal basis in an infinite dimensional Hilbert space

or you consider any separable Hamel basis and extract an orthonormal basis aftermath by the of Gram-schimdt process.

Whereas the first question is more technical. In fact it springs from Bessel-Parseval identity we have $$\color{blue}{x=\sum_{n = 1}^{\infty}\langle x,e_n \rangle e_n~~~~~and ~~~~~ \|x\|^2 = \sum_{n = 1}^{\infty}|\langle x,e_n \rangle|^2} $$

Employing Cauchy-Schwarzt inequalities it follows that,

$$\color{black}{|f(x)| = \left|\sum_{n = 1}^{\infty}\frac{1}{3^n}\langle x,e_n \rangle\right| \le \left(\sum_{n = 1}^{\infty}\frac{1}{3^{2n}}\right)^{1/2} \left(\sum_{n = 1}^{\infty}|\langle x,e_n \rangle|^2\right)^{1/2} = \frac{1}{2\sqrt{2}}\|x\|}$$

Now let $a= \sum_{n=1}^\infty \frac{2\sqrt2}{3^{n}}e_n $ from Bessel-Parseval identity aigain we have $$\|a\|^2 = \sum_{n=1}^\infty \frac{8}{3^{2n}} = 1 ~~~and~~~\langle a,e_n\rangle= \frac{2\sqrt2}{3^n}$$

Then, $$\color{red}{ f(a)=\sum_{n=1}^\infty \frac{2\sqrt2}{3^{2n}} =\frac{2\sqrt{2}}8= \frac{1}{2\sqrt{2}}}$$ Hence by definition of the operator norm, we have

$$\frac{1}{2\sqrt{2}} \ge \|f\|=\sup\{\|f(x)\|: \|x\|^2=1\}\ge f(a)= \frac{1}{2\sqrt{2}}$$

therefore,

$$\color{red}{\|f\|= \frac{1}{2\sqrt{2}}}$$

Answer 4

$f$ is a continuous linear functional given by the Riesz Representation as $$ f(x) = \langle x,\sum_{n=1}^{\infty}\frac{1}{3^n}e_n\rangle = \langle x,y\rangle. $$ The norm of such a functional is $$ \|f\|=\|y\|=\left(\sum_{n=1}^{\infty}\frac{1}{3^{2n}}\right)^{1/2} = \frac{1}{3}\left(\sum_{n=0}^{\infty}\frac{1}{9^n}\right)^{1/2}=\frac{1}{3}\left(\frac{1}{1-1/9}\right)^{1/2}=\frac{1}{2\sqrt{2}}. $$