Show the map $f(x)=\|Lx\|$ is a functional on the normed space $X$

Question

I am trying to solve something but I think it might not be possible... I'm looking for either a solution, or confirmation of my suspicion

Given a normed space $(X,\| \cdot \|)$ and bounded linear functional $L:X \to X$, I want to be able to show that the function $f: X \to \mathbb{R}$ defined by $$ f(x) = \|L\| $$ is linear. In other words, I want to show that for all $\alpha, \beta \in \mathbb{R}$ and $x,y \in X$, $$ f(\alpha x + \beta y) = \alpha f(x) + \beta f(y) $$

My solution attempt:

I get stuck almost immediately because I cannot avoid the "$\leq$", or the absolute values, in the following argument: \begin{alignat*}{2} f(\alpha x + \beta y) &= \|T(\alpha x + \beta y)\| \\ &= \| \alpha Tx + \beta Ty \|&& \mbox{by linearity of } T \\ &\leq \| \alpha Tx \| + \| \beta Ty \|&& \mbox{by Triangle Inequality of } \| \cdot \| \\ &= | \alpha | f(x) + | \beta | f(y) &\quad\;&\mbox{by Homogeneity of } \| \cdot \| \end{alignat*}

Answer 1

Let $L$ be the identity. The norm is not linear, that's easy to show.