Consider $L^p(\mathbb{R})$ where $0<p<1$.
Why there are no bounded linear functional on $L^p(\mathbb{R})$? i.e. If $l$ is linear functional $l:L^p(\mathbb{R})\to \mathbb{C}$ such that $|l(f)|\leq M|f|_{p}$ for all $f$ and some $M>0$, then $l=0$. Why? Hint: Let $F(x)=l(\mathcal{X}_{[0,x]})$, and consider $F(x)-F(y)$
This question surely was already answered here. In the case $0 <p <1$ the usual definition of the $L^p$-norm is not a norm, because the $\Delta$-inequality isn't satisfied. If $0< p <1$ the space $L^p(\mathbb{R})$ is only a metric space with invariant metric given by $$\|f\|_p := \int |f|^p d \lambda.$$ If $l$ is a continuous functional, we know that $|l(f)| < 1$ for $f \in B_\delta(0)$ and some $\delta >0$. By rescaling and using linearity, we see that $$|l(f)| \le M \|f\|_p^{1/p}$$ with $M= \delta^{-1}$. Now decompose $[a,b]$ in $n$ disjoint intervals $I_1,\ldots,I_n$ with lenght $(b-a)/n$. Then we have $$|l(1_{[a,b]})| \le \sum_{k=1}^n |l(1_{I_i})| \le nM \|l(1_{I_i})\|_p = (b-a)^{1/p} n^{1-1/p}.$$ Since $0<p<1$, we see that $1-1/p$ is negative. Thus, letting $n \rightarrow \infty$, we get $l(1_{[a,b]}) =0$. Since all linear combinations of such intervals are dense in $L^p$ (also for $0<p<1$), we see that $l$ is trivial.
Suppose $\phi\in (L_p(\mathbb{R},\lambda))^*$ where $\lambda$ is the Lebesgue measure on $\mathscr{B}(\mathbb{R})$. If $\phi\neq0$, then there is $f\in L_p$ such that $\phi(f)\neq0$. Then either $\phi(f_+)>0$ or $\phi(f_-)>0$. This means that without loss of generality, we may assume that $f\geq0$ and $\phi(f)\leq1$. Define $$F(x)=\int^x_{-\infty}f^p(t)\,dt$$ $F$ is a continuous function such that $\lim_{x\rightarrow-\infty}F(x)=0$ and $F(+\infty):=\lim_{x\rightarrow\infty}F(x)=\int f^p(t)\,dt$. Hence there is a point $x_1$ such that $F(x_1)=\frac12 F(+\infty)$. Define \begin{align} g_1 &=f\mathbb{1}_{(-\infty,x_1]},\qquad g_2 &=f\mathbb{1}_{(x_1,\infty)} \end{align} Clearly $f^p=g^p_1+g^p_2$. Since $\phi(g_1)+\phi(g_2)=\phi(f)\geq1$, then either $\phi(g_1)\geq \frac12$ or $\phi(g_2)\geq\frac12$. Choose $g\in \{g_1,g_2\}$ such that $\phi(g)\geq\frac12$, and define $f_1=2g$. Observe that $\phi(f_1)\geq1$ and $$d(f_1,0)=2^p\int g=2^{p-1}\int f$$ Applying the same argument to $f_1$ in place of $f$, one obtains a function $f_2$ such that $\phi(f_2)\geq1$ and $$d(f_2,0)=2^{p-1}\int f^p_1=2^{n(p-1)}\int f^p$$ Continue this way add infinitum, we obtain a sequence $(f_n:n\in\mathbb{Z}_+)$ such that
Since $0<p<1$, $d(f_n,0)\xrightarrow{n\rightarrow\infty}0$. The continuity of $\phi$ then would imply that $\phi(f_n)\xrightarrow{n\rightarrow\infty}0$; this however contradicts (2). Consequently $\phi=0$.
Here is a solution which may be what the writer of the exercise was intending in the hint:
If $y \geq x$, then by linearity of $l$, we have $$|F(y) - F(x)| = |l(\chi_{[x,y]})| \leq M|x - y|^{\frac{1}{p}} $$
If $ x \geq y$ then similar computations apply and we can see $F(x)$ is Hölder continuous with exponent $\frac{1}{p} > 1$. Therefore $F$ is constant and $ l(\chi_{[x,y]}) = 0 $ for all intervals $[x,y]$. We then conclude by density of step functions in $L^p$ (which holds even for $0 < p < 1$).
