Show that $L^{p}(\mathbb{R}^d)\not\subset L^{\infty}(\mathbb{R}^d)$ for all $p\in[1,\infty).$
My attempt at this problem:
I define the following function for $x\in\mathbb{R}^d,$ $$f_a(x)=\begin{cases}\vert x\vert^{a}&\text{if }0<\vert x\vert<1\\0&\text{otherwise. }\end{cases}$$ I am able to prove that this function is in $L^{1}(\mathbb{R}^d)$ if and only if $a>-d.$ So I define another function $$f(x)=\begin{cases}\vert x\vert^{-d/(p+1)}&\text{if }0<\vert x\vert<1\\0&\text{otherwise. }\end{cases}$$ Then since $-dp/(p+1)>-d,$ we have $$\large\|f\|_p^p=\int_{\mathbb{R}^d}\vert x\vert^{-dp/(p+1)}\,dx<\infty,$$ so that $f\in L^{p}(\mathbb{R}^d).$
My difficulty is with showing that $f\notin L^{\infty}(\mathbb{R}^d),$ since for any $M>0$ we have that $f>M$ on a set not of measure zero. However we can choose a larger $M$, and then the measure of the set where $f$ is not bounded decreases.
I my reasoning above correct, and $f$ actually works, or am I missundertanding the notion of the $\|\|_{\infty}$ norm?
Appreciate any feedback.
Thank you for your time.
If $f \in L^{\infty}$ the there exists a constant $M \in (0,\infty)$ such that $|f(x)| \leq M$ almost everywhere. This gives $|x|^{-d/(p+1)} \leq M$ for almost all points $x$ in the region $0<|x|<1$. Actually, since LHS is continuous the inequality must hold for all $x$ in the region $0<|x|<1$. Can you get a contradiction from here by taking $x$ close to the origin?