Let $X$ be a Hilbert space and $\{e_1,\dots, e_n,\dots \}$ be an orthonormal set. Let $(\alpha_m)$ be a sequence in $\mathbb{R}$. Show that \begin{align*} &Lu = \sum_{m=1}^{\infty}\alpha_m\langle u,e_m \rangle e_m& \end{align*}
defines a bounded linear map $L$ on $X$ if and only if $K=\sup_{ m=1}^{\infty} |\alpha_m| <\infty.$
My attempt: Let $K=\sup_{ m=1}^{\infty} |\alpha_m| < \infty$, and we show that $L$ is bounded and linear: To show that $L$ is linear, let $a, b$ be scarlas and $u, v$ be vectors in $X$, then \begin{align*} &L(au+bv)=\sum_{m=1}^{\infty}\alpha_m\langle au +bv,e_m \rangle e_m&\\ &=a\sum_{m=1}^{\infty}\alpha_m\langle u ,e_m \rangle e_m+b\sum_{m=1}^{\infty}\alpha_m\langle v ,e_m \rangle e_m&\\ &=aLu+bLv& \end{align*} This shows the linearity of $T$.
To show the boundedness, \begin{align*} &\|Lu\| = \|\sum_{m=1}^{\infty}\alpha_m\langle u,e_m \rangle e_m \|&\\ &= \|\sum_{m=1}^{\infty}\alpha_m\langle u,e_m \rangle e_m \|&\\ &\leq \sum_{m=1}^{\infty}|\alpha_m| |\langle u,e_m \rangle| \| e_m \|&\\ &= \sum_{m=1}^{\infty}|\alpha_m| |\langle u,e_m \rangle|&\\ &\leq|K| \sum_{m=1}^{\infty} |\langle u,e_m \rangle|&\\ \end{align*} From here I am not sure how to proced to show the boundedness. And also I am not sure how to show the converse. Any feedback will be appreciated. Thanks.
As already pointed out the result is false as stated. The correct statement is $L$ is a bounded operator iff $\sup_m |\alpha_m| <\infty$.
For a proof you have to recall that $\|v\|^{2} \geq \sum |\langle v, e_n \rangle|^{2}$ for any vector $v$ and any orthonormal sequence $(e_n)$.
From this it follows that $\|Lv\|^{2} \leq M^{2}\|v\|^{2}$ where $M=\sup_m |\alpha_m| <\infty$. Hence $L$ is a bounded operator when $M <\infty$. If $M=\infty$ then $\|L\| \geq \|Le_k\|= |\alpha_k| \to \infty$ so $L$ is not bounded.