Compute the operator-norm $L \in B(V,V)$ Let $V = C_\mathbb{R}[0,1]$, define $L \in L(V,V)$ as $(Lf)(x) = xf(x)$. Equip $V$ with the $L^2$-norm. I want to compute the operator norm 
$$
\|L\|_{B(V,V)} := \sup\{\|Lf\|_{L^2} : \|f\|_{L^2} \leq 1\}
$$
And I have to show that for any $0 \neq f \in V$ we have have $\|Lf\|_{L^2} < \|L\|_{B(V,V)} \|f\|_{L^2}$.
What I have tried so far. To show $L \in B(V,V)$, I computed: 
$$
\|Lf\|_{L_2}^2 = \int_0^1 |xf(x)|^2dx = \int_0^1 |x|^2|f(x)|^2dx
\leq \int_0^1 |f(x)|^2dx = \|f\|_{L_2}^2. 
$$
Thus, we have $\|Lf\|_{L_2} \leq k \|f\|_{L_2}$ for $k=1$, and $L$ is bounded. It also follows that $\|L\| \leq 1$.
I expected that we can show somehow that $\|L\| \geq 1$ by showing that there is some sequence $g_n \in C_\mathbb{R}[0,1]$ for which the norm converges to $1$, but I didn't manage to do this, I'm not sure if that's correct, someone has any ideas on how to do this?
 A: $\newcommand{\a}{\alpha}\newcommand{\b}{\beta}$
$\newcommand{\norm}[1]{\left\lVert#1\right\rVert}$
Consider functions of the form $f(x) = \a \dfrac{\sqrt{x - a}}{\sqrt{1 - \b}}$ when $x \geq \b$, otherwise $f(x) = 0$.
By calculation, $\norm{f} = \dfrac{\a^2}{1 - \b}\dfrac{(\b - 1)^2}{2}$. To set this to $1$, use $\a^2 = (1 - \b)\dfrac{2}{(\b - 1)^2}$.
Also $\norm{Lf} = \dfrac{\a^2}{1 - \b}\dfrac{\b^4 - 4\b + 3}{12}$. With the above restriction of choosing $\a, \b$ such that $\norm{f} = 1$, the expression for the norm is $\norm{Lf} = \dfrac{\b^4 - 4\b + 3}{12}\dfrac{2}{(\b - 1)^2} \to 1$ as $\b \to 1$.
Now, formally, we need to pick a series $f_n$ in your function space such that $\norm{f_n} = 1$ and $\norm{Lf_n} \to 1$. To do that set $f_n$ to be the $f$ with $\b = 1 - 1/n$, and $\a$ set as above so that $\norm{f_n} = 1$. Now, since $\b \to 1$, as $n \to \infty$, $\norm{Lf_n} \to 1$, as required, and so the operator 
Note: This is definitely not the nicest function sequence, but yes, it (probably?) works.
A: $\newcommand{\norm}[1]{\left\lVert#1\right\rVert}$
To reduce calculations, the following answer is probably way more tractable.
Pick $f_n(x) = \sqrt{2n + 1}x^n$. This "concentrates mass" near $1$. Also, trivially, $\norm{f_n} = 1$.
Also $\norm{Lf_n} = \displaystyle\int_{0}^{1} (2n + 1)x^{2n + 2} = \dfrac{2n + 1}{2n + 3} \to 1$ as $n \to \infty$ as required.
