Let the operator $T : C[0, 1] → C[0, 1]$ be defined by
$$(T f)(x) = x^2f(x)$$ for x ∈ [0, 1]. If we equip C[0, 1] with the norm
$$||f||_{L^p[0,1]} =(\int^ 1_0|f(x)|^p dx)^{1/p}$$ for $1 ≤ p < ∞$, show that $T$ is linear, bounded, and find its operator norm $||T||$.
Proof idea: Linearity follows simply from the definition of Tf. (no questions here)
Boundedness (thus continuity):
$||Tf||^p = ||x^2 f(x)||^p = \int_0^1 |x^2f(x)|^p dx =\int_0^1 |x^{2p}||f(x)|^p dx\leq
\int_0^1 |f(x)|^p$ since $x^{2p}\in [0,1]$.
So $||Tf||^p = 1 ||f||^p_{L^p[0,1]}<\infty$ so $||Tf||$ is bounded thus continuous.
$||T||=sup\frac{||Tf||}{||f||}\leq sup\frac {||f||}{||f||}=1$.
Let $f_1=1 \in C[0,1]$ then $||f_1||=1$ and $||T||=sup\frac{||Tf||}{||f||}\geq \frac {||Tf_1||}{||f_1||}=1/1=1$. So $||T||=1$.
I am uncertain about it. Is this the norm? also is my proof of boundedness correct?
Thanks and regards,
Best Answer
Your proof that $T$ is bounded is correct.
Your mistake comes when you say the following:
This is not true, as $$\|f_1\|=\left(\int_{0}^1 |x^2|^p dx\right)^\frac1p = \left(\frac{1}{2p+1}\right)^\frac1p\neq 1$$
You are, however, correct that the norm of $T$ is indeed $1$, but it is slightly harder to show this than you hoped.
In fact, you will not find a single non-zero continuous function for which $\|Tf\|=\|f\|$, but that doesn't mean the norm isn't $1$. In fact, you can also prove the norm is $1$ by finding a sequence of functions $f_n$ such that $\frac{\|Tf_n\|}{\|f_n\|}$ approaches $1$ as $n$ becomes large.
Hint:
In particular, think about what happens to $\frac{\|Tg_n\|}{\|g_n\|}$ when
$$g_n(x)=\begin{cases}0&x<1-\frac{1}{n}\\1&x\geq1-\frac1n.\end{cases}$$
Note: of course $g_n$ is not continuous, but thinking about what happens to the function $g_n$ may still give valuable insight. Also, even if it is not continuous, you can create a function $f_n$ that is very similar to $g_n$, but also continuous.