Find the norm of the linear operator $f(x) = \int_0^2x(t)dt$

Question

a) $f: L^1(0,3) \rightarrow \mathbb{R}$

b) $f: C[0,3] \rightarrow \mathbb{R}$

for part a I got $\|f\| = 1$ because $\|f(x)\|=|\int_0^2x(t)dt| \leq \int_0^2|x(t)|dt \leq\int_0^3|x(t)|dt = \|x(t)\|_1$ so $\|f\|=1$

for b, I think its similar: $\|f(x)\|=|\int_0^2x(t)dt| \leq \int_0^2|x(t)|dt \leq\int_0^3|x(t)|dt = \|x(t)\|_1$ and I have a theorem that says there is some c>0 s.t. $\|x(t)\|_1 \leq c\|x(t)\|_{max}$ so by taking the infinum of all such c, we get that $\|f\| = 0$.

Are these correct? I am particularly worried about my answer for b

Answer 1

b) is not correct. There is some $c>0$ such that $\|x\|_{1}\leq c\|x\|_{\infty}$, then $\|f\|\leq c$ for this $c$, it does not mean that for every $c>0$, $\|f\|\leq c$.

Rather, the question needs the exact norm of the operator. Actually one has $|f(x)|=\left|\displaystyle\int_{0}^{2}x(t)dt\right|\leq\displaystyle\int_{0}^{2}|x(t)|dt\leq\|x\|_{\infty}\int_{0}^{2}1dt=2\|x\|_{\infty}$.

Now take $x(t)=1$ for $0\leq t\leq 2$ and $x(t)=-t+3$ for $2\leq t\leq 3$, then $\|x\|_{\infty}=1$ and hence $|f(x)|=2$, finally $\|f\|=2$.

Answer 2

For part a), you have only shown $\|f\| \le 1$. But it should not be hard to find some $x$ such that $|f(x)| = \|x\|$.

For part b), note $|f(x)| \le \int_0^2 |x(t)|\,dt \le 2 \|x\|_{\max}$, so $\|f\| \le 2$. If you take some $x \in C[0,3]$ with $x(t)=1$ for $t \in [0,2]$ and $\|x\|_{\max}=1$, then $|f(x)|=2=2\|x\|_{\max}$, so $\|f\|=2$.