finding a function in $L^{p}(0,1)$ that is not $L^{q}(0,1)$ where $p < q$

Question

Let $ p < q$. I have shown that $L^{q}(0,1)\subseteq L^{p}(0,1)$ but now I want to show that $L^{q}(0,1)$ is a proper subspace of $L^{q}(0,1)$.

My initial idea was to set $f(x):=\frac{1}{x^{\frac{1}{2p}}}$. It is clear that

$\vert \vert f \vert \vert_{L^{p}}^{p}=\int_{0}^{1}\vert \frac{1}{x^{\frac{1}{2p}}}\vert ^{p}dx=\int_{0}^{1} \frac{1}{\sqrt{x}}dx<\infty$ which is great but in order for $f \notin L^{q}(0,1)$ my $q$ would have to be $\geq 2p$

Any hints?

Answer 1

Hint: recall that the integral

$$ \int_0^1 \frac{1}{t^s}dt $$

is finite if and only if $s < 1$.

Hence, we'd like to find $\alpha > 0$ such that $\alpha p < 1$ but $\alpha q \geq 1$. Since $p < q$, then $\frac{p}{q} < 1$ and $\frac{q}{q} = 1$. Thus we can choose $\alpha = \frac{1}{q}$.