Which of the mappings $x\rightarrow F(x)$ are linear and continuous functionals over $L^2(0,1)$? .
$$(i)\qquad F(x)=\int_0^1\frac{x(t)}{\sqrt{t}}dt$$ $$(ii)\qquad F(x)=\left(\int_0^1x(t)^2dt\right)^{1/2}$$ $$(iii)\qquad F(x)=\int_0^1\int^t_0x(s)dsdt$$ my ideas:
for $(i)$: i think this mapping is probably not bounded and therefore not continuous, but i don't know how to estimate F(x) to get a violation of $||Fx||<M||x||$
for $(ii)$: this mapping is not linear, at least i don't sea how $F(\alpha x+\beta y)=\alpha F(x)+\beta F(y)$ should work here
for $(iii)$ this mapping seems to be a linear and bounded (continuous) functional, but im not quite sure how to show the boundedness. maybe $F(x)=\int_0^1\int^t_0x(s)dsdt\leq\sup_t\int^1_0\int^t_0x(s)ds...$
Then i have to find the riesz representation for the linear and continuous functionals. so the $f\in L^2(0,1)$ such that $F(x)=\left\langle x,f\right\rangle \forall x\in L^2(0,1)$ an calculate the Norm of $L^2(0,1)'$.. https://en.wikipedia.org/wiki/Riesz_representation_theorem
Maybe you guys can tell if my intuition is right and give me some tips on how to proceed.
For (i) note that the function $f(t) = { 1 \over \sqrt{t}}$ is not in $L^2[0,1]$ (because of the behaviour at $t=0$), so this suggests that $F$ is not continuous, and that we should look for a counterexample that exploits the behaviour at $t=0$.
Let $x_n(t) = {1 \over 2(1-{1 \over \sqrt{n}})}1_{[{1 \over n},1]}{1 \over \sqrt{t}}$ and note that $\|x_n\|=1$ and $F(x_n) = {1 \over 2(1-{1 \over \sqrt{n}})} \ln n$.
For (ii), just note that $F (x \mapsto 1) = F (x \mapsto -1) >0$, hence $F$ cannot be linear.
For (iii), write $F(x) = \int_0^1 \int_0^1 x(s) \cdot 1_{[0,t]}(s) ds dt$, it is clear that $F$ is linear. Also, Cauchy Schwarz shows that $\int_0^1 |x(s) \cdot 1_{[0,t]}(s)| ds \le \|x\| \| 1_{[0,t]} \| = \sqrt{t} \|x\|$, and so $|F(x)| \le {2 \over 3} \|x\|$.
By interchanging the order of integration in $F(x) = \int_0^1 \int_0^1 x(s) \cdot 1_{[0,t]}(s) ds dt$ you can find and compute an explicit formula for some $\phi$ such that $F(x) = \int_0^1 x(s) \phi(s) ds$.