Show that $f \in c_0^*$ and $||f||=\sum_{j=1}^{\infty} \frac{1}{j!}$

Question

Let $$\begin{eqnarray} f: c_0 & \to & \mathbb{R}\\ (x_i)_1^{\infty} & \to & \displaystyle \sum_{j=1}^{\infty} \frac{x_j}{j!}\\ \end{eqnarray}$$ Show that $f \in c_0^*$ and $||f||=\sum_{j=1}^{\infty} \frac{1}{j!}$.

I can show that $f \in c_0^*$ but can't show the restant. $c_0=(x_i)_1^{\infty} \ ; \ x_i \to 0, \ n \to \infty$

Answer 1

We have $$||f(x_i)||=\left|\sum_{j=1}^{\infty} \frac{x_j}{j!}\right|\leq \sum_{j=1}^{\infty} \frac{|x_j|}{j!} \leq ||(x_i)||_\infty \sum_{j=1}^{\infty} \frac{1}{j!} $$ so we have $$||f||\leq \sum_{j=1}^{\infty} \frac{1}{j!}$$ Now let $(x^k_i)$ defined by $x_1=\cdots,x_k=1$ and $x_i=0$ if $i>k$ so $||(x^k_i)||_\infty=1$ and then $$\frac{|f(x^k_i)|}{||(x^k_i)||_\infty}=\sum_{j=1}^{k} \frac{1}{j!}\leq ||f||$$ and pass to the limit $k\to \infty$.

Answer 2

Hint: $$ \vert f(x)\vert\leq\sum_{j=1}^\infty\left\vert\frac{x_j}{j!}\right\vert\leq\sup_j\vert x_j\vert\sum_{j=1}^\infty\frac{1}{j!} $$

Then, consider sequences of the form $(x_j)=(1,1,1,\ldots,0,0,\ldots)$, and send the ones off to infinity; this sequence of sequences will have $\vert f(x)\vert\rightarrow\sum_{j=1}^\infty 1/j!$

Answer 3

Hint: Suppose $\mathbf{x}:=(x_n)_{n=1}^{\infty}$ is in $c_0$. Then $$ \lvert x_n\rvert\leq\sup_{m\in\mathbb{N}}\,\lvert x_m\rvert=\|\mathbf{x}\|\tag{1} $$ for all $n$. So, if you start with $$ \left\lvert\sum_{n=1}^{\infty}\frac{x_n}{n!}\right\rvert\leq\sum_{n=1}^{\infty}\frac{\lvert x_n\rvert}{n!}, $$ and apply (1), you immediately get that $$ \|f\|\leq\sum_{n=1}^{\infty}\frac{1}{n!}. $$ So, what remains is to show that this is the best upper bound. To do this, it suffices to consider only $\mathbf{x}\in c_0$ such that $\|\mathbf{x}\|=1$. Try thinking about the sequence of elements $\mathbf{x}^{(j)}\in c_0$ given by $$ x_n^{(j)}=\begin{cases}1 & \text{if $n\leq j$}\\ 0 &\text{else}\end{cases}, $$ where $x_n^{(j)}$ denotes the $n$th element of $\mathbf{x}^{(j)}$.