Calculating Operator Norms of an Integral Operator

Question

Let $K= [0,a]$ and define norms in $C(K)$, the maximum norm $||f||_{0}$ and weighted max norm $||f||_{1} = max|f(x)|e^{-ax}$. Calculate the operator norms $||T||_{0}$ and $||T||_{1}$ for the linear operator $ Tf(x) = \int^{x}_{0} tf(t)dt$.

Answer 1

For $\|T\|_0$ :

Let $f \in C(K)$, $\|f\|_0 = 1$

$$\left| Tf(x) \right| = \left| \int_0^x t f(t) dt \right| \leq \int_0^x t |f(t)| dt \leq \int_0^x t dt = \frac{x^2}{2}$$

So $\|Tf\|_0 \leq \frac{a^2}{2}$.

We also have that for $f(x) = 1$, $\|Tf\|_0 = \frac{a^2}{2}$

Hence $\|T\|_0 = \frac{a^2}{2}$

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For $\|T\|_1$ :

Let $f \in C(K)$, $\|f\|_1 = 1$

$$\left| Tf(x) \right|e^{-ax} = \left| \int_0^x t f(t) dt \right|e^{-ax} \leq e^{-ax}\int_0^x t e^{at}\underbrace{|f(t)|e^{-at}}_{\leq 1}dt $$

$$\leq e^{-ax}\int_0^x te^{at} dt = e^{-ax} \left( \frac{e^{ax} (ax-1) - 1}{a^2} \right)$$

$$ \leq 1- \frac{1}{a^2}( 1 - e^{-a^2} )$$

So $\|T\|_1 \leq 1- \frac{1}{a^2}( 1 - e^{-a^2} )$

Now take $f(x) = e^{ax}$, $\|f\|_1 =1$

$$|Tf(x)|e^{-ax} = e^{-ax} \int_0^x t e^{at} dt = e^{-ax} \left( \frac{e^{ax} (ax-1) - 1}{a^2} \right)$$

$$= \frac{ax-1}{a^2} - \frac{1}{a^2}e^{-ax}$$

Now $Tf$ is increasing, so $|Tf(x)|e^{-ax}$ has a maximum at $x=a$, hence

$$\|Tf\|_1 = 1- \frac{1}{a^2}( 1 - e^{-a^2} )$$

Hence $\|T\|_1 \geq 1- \frac{1}{a^2}( 1 - e^{-a^2} )$

In conclusion, we have the equality :

$$\|T\|_1 \geq 1- \frac{1}{a^2}( 1 - e^{-a^2} )$$