Calculating norm of an operator Let $X=C\left([0,1]\right)$ be equipped with norm 
$\|f\|:=\sqrt{\int\limits_0^1 |f(t)|^2 dt \ + \sum\limits_{n=1}^\infty 2^{-n} |f(x_n)|^2}$,
where $\{x_n\}_{n=1}^\infty \subset[0,1]$, s. th. $x_n\neq x_m$ if $n\neq m$.   
What is the norm of the evaluation operator 
$\Psi_x: X\rightarrow \mathbb{C}, \ \Psi_x f= f(x)$ ?
 A: To study this, we can study separately the cases:


*

*$x\neq x_n$ for all $n$ and is not an accumulation point of $(x_n)_n$

*$x\neq x_n$ for all $n$ but is an accumulation point of $(x_n)_n$

*$x=x_n$ for some $n$


In the first case, the operator is unbounded. Indeed, there is $\epsilon>0$ s.t. for all $n$, $\vert x-x_n\vert >\epsilon$. For all $n\in\mathbb{N}$, we can find a continuous function $f_n$ s.t. 


*

*$f_n(z)=0$ if $\vert z-x\vert \geq \epsilon/2$

*$f_n(x)=n$

*$\Vert f_n\Vert_{L^2}=\Vert f\Vert\leq 1$


which shows the operator is unbounded. 
In the second case, the operator is again unbounded. Let $n\in\mathbb{N}$, $m_n=\min_{j\in\{1,..,n\}}\vert x-x_j\vert>0$. We can find a continuous function $f_n$ s.t. 


*

*$f_n(z)=0$ if $\vert z-x\vert \geq m_n/2$

*$f_n(x)=2^{n/4}$

*$0\leq f_n(z)\leq f_n(x)$ for all $z$

*$\Vert f_n\Vert_{L^2}\leq 1/2$, which implies 
$$
\Vert f_n\Vert^2\leq \Vert f_n\Vert_{L^2}^2+\sum_{j\geq n+1} 2^{-j}\vert f_n(x_j)\vert^2\leq 1/4+2^{-n}2^{n/2}\leq 1
$$


All this shows the operator is unbounded. 
In the third case, the norm of $\Psi_x(f)$ is $2^{n/2}$. Indeed, let $f$ s.t. $\Vert f\Vert \leq 1$. Then 
$$
2^{-n} \vert f(x_n)\vert^2 \leq 1\Rightarrow \vert \Psi_{x_n}(f)\vert=\vert f(x_n)\vert\leq 2^{n/2}
$$
which shows the norm of $\Psi_x(f)$ is bounded by $2^{n/2}$. To show equality, consider $g(z)=2^{n/2}1_x(z)$. This function is in $L^2([0,1])$ and $\Vert g\Vert =1$. The idea of the proof is to approach $g$ by an "apropriate" sequence of continuous functions. An idea to do this, is the following. Let $k\in\mathbb{N}$, $m_k=\min_{j\in\{1,..,k\}\setminus\{n\}}\vert x-x_j\vert>0$. We can define a continuous function $f_k$ s.t. 


*

*$\Vert f_k\Vert_{L^2}\to 0$ as $k\to\infty$

*$f_k(z)=0$ if $\vert z-x\vert \geq m_k/2$

*$0\leq f_n(z)\leq f_n(x)$ for all $z$

*$f_k(x_n)=\sqrt{2^n\left(\frac{1-\Vert f_k\Vert_{L^2}^2}{1+2^{n-k}}\right)}\to 2^{n/2}$. This also implies 
\begin{eqnarray}
\Vert f_k\Vert^2 & \leq & \Vert f_k\Vert_{L^2}^2+2^{-n}\vert f_k(x_n)\vert^2+\sum_{j\geq k+1} 2^{-j}\vert f_k(x_j)\vert^2\\
& \leq & \Vert f_k\Vert_{L^2}^2+(2^{-n}+2^{-k})\vert f_k(x_n)\vert^2\\
& \leq & 1
\end{eqnarray}


which shows the norm of $\Psi_x(f)$ is $2^{n/2}$
