Finding the norm of the given operator

Question

Let's define an operator: $$T: L^p([0, 1]) \to L^p([0, 1]),$$ $$(Tf)(x) = xf(x).$$ Where $L^p = \big\{ f: \int \limits_{0}^{1}|f(x)|^p \mbox{d}x < \infty \big\}$.
We are to find the norm of the given operator.
Firstly I managed to show that: $$|| Tf ||^p = \int\limits_0^{1} |x|^p|f(x)|^p \mbox{d}x \le \int \limits_{0}^{1} |f(x)|^p \sup_{x \in[0, 1]}|x|^p \mbox{d}x = ||f||^p.$$ Thus we suspect that $||T|| = 1$.
Now we just need to find an element from $L^p([0, 1])$ such that it's norm is equal to $1$. What should the element look like?

Answer 1

If you want to find some $f$ such that $\|Tf\|/\|f\| = 1$, take a look at the inequality you found which yielded $\|Tf\| \leq \|f\|$, and see if you can find a case which makes it an equality. You determined this by making the estimate:

$$ |x|^p \leq \sup \limits_{x \in [0,1]} |x|^p $$

This is not a very sharp estimate. In fact, for any $t \in (0,1)$, we could rewrite your estimate to get:

$$ \|Tf\|^p = \int \limits_0^1 |f(x)|^p |x|^p dx \leq t^p \int \limits_0^t |f(x)|^p dx + \int \limits_t^1 |f(x)|^p dx \\ \leq \int \limits_0^1 |f(x)|^p dx = \|f\|^p $$

But here, this last inequality is only an equality if $f$ is zero on $[0,t]$. So, it seems like the only function which gives you an equality is the zero function, but this doesn't help you find the norm.

Instead, you could try finding a sequence of functions $(f_n)_n$ such that $\|Tf_n\|/\|f_n\| \rightarrow 1$: this also would let you prove that $\|T\|=1$. But how to find this sequence? Again, we look at the estimate you made to get your inequality $\|Tf\| \leq \|f\|$. What functions could you pick to minimize the effect of your estimate? Since you used the estimate with $\sup_{x \in [0,1]} |x|^p$, it makes sense to look for functions which are "concentrated" near 1, since this is where the difference between $|x|^p$ and $\sup_{x \in [0,1]} |x|^p$ is the smallest. In particular, as it's been pointed out in the comments, you could try the characteristic function on $[1-\frac{1}{n}, 1]$.