For simplicity, I write $L^p$ instead of $L^p([0,1])$. For $p<q$, we have $\lVert\cdot\rVert_p\le\lVert\cdot\rVert_q$ and hence $L^q\subseteq L^p$. Here $L^2\subseteq L^p$ for all $p<2$.
Let us take a non-zero $g\in L^2$ and define $S=\{f\in L^2: \langle g,f\rangle_{L^2}=0\}$. Then $S^\perp=\text{Span}(g)\ne0$, hence $S$ is not dense in $L^2$. I want to examine whether $S$ is dense in $L^p$ or not.
Let us take $p<2$, then S is dense in $L^p$ iff the only linear functional on $L^p$ which vanishes on $S$ is the zero linear functional. We know that $(L^p)^*=L^q$ where $p,q$ are conjugate. As $p<2$, $q>2$. Let $h\in L^q$ and $\int hf\ dm=0$ for all $f\in S$.
Now if $\int gh\ dm=0$ i.e. $\overline{h}\in S$, we have $\int h\overline{h}\ dm=0\implies h=0$.
Therefore, $S$ is dense in $L^p$ for every $p<2$ if and only if there is non-zero $g\in L^2$ such that $\int gh\ dm=0$ for all $h\in L^q$ for every $q>2$.
But is it possible to construct such $g$? Can anyone help me in this regard? Is there any alternate way-out to solve the problem?
Thanks for your help in advance.
The function $g(x)=x^{-1/2}(-\log (x/2))^{-1}$ belongs to $L^2(0,1)$ and does not belong to $L^q(0,1)$ for $q>2.$ Let $$V=\left\{f\in L^2(0,1)\,:\langle f,g\rangle =\, \int\limits_0^1 f(x)\overline{g(x)}\,dx =0\right \}$$ Then $V$ is not dense in $L^2.$ Assume for a contradiction that $V$ is not dense in $L^p$ for some $1\le p<2.$ Then there exists a bounded linear functional on $L^p$ which vanishes on $V.$ By duality between $L^p$ and $L^q,$ $q=p/(p-1),$ there is a function $0\neq h\in L^q$ such that $$\int\limits_0^1 f(x)\overline{h(x)}\,dx =0,\quad f\in V$$ We have $h\in L^2.$ Let $$ u=\|g\|^2h-\langle h,g\rangle g$$ Then $u\in V,$ because $\langle u,g\rangle =0,$ and $$\int\limits_0^1 f(x)\overline{u(x)}\,dx =0,\quad f\in V$$ Plugging in $f=u$ gives $\langle u,u\rangle =0,$ i.e. $u=0.$ Therefore $h=a g,$ for a nonzero constant $a,$ a contradiction, as $g\notin L^q.$