Inclusions of $L^1[a,b]$, $L^2[a,b]$ and $L^3[a,b]$

Question

Define

$$L^p[a,b]=\{f:[a,b]\to\mathbb{R}\,\big| \int_a^b|f(x)|^p\text{d}x < \infty \}.$$

I want to show the followings

$1$. If $g\in L^1[a,b]$ and $g\in L^2[a,b]$ then $g\in L^3[a,b]$.

$2$. $L^2[a,b]\subset L^3[a,b].$

$3$. Give an example of a function $g$ that $g\in L^3[a,b]$ and $g \notin L^2[a,b]$.

There is a hint which says if $p,q\ge1$ and $\frac{1}{p}+\frac{1}{q}=1$ then we have

$$\int_a^b |f(x)g(x)|\,\text{d}x\le \Big(\int_a^b |f(x)|^p\,\text{d}x\Big)^{\frac{1}{p}} \Big(\int_a^b |g(x)|^q\,\text{d}x\Big)^{\frac{1}{q}}.$$

Does this inequality have a name? and where can I find a proof for it? (I don't know much about functional analysis so please refer me to a simple proof).

I used the inequality for $g(x)=1$ to obtain

$$\int_a^b |f(x)|\,\text{d}x\le \Big(\int_a^b |f(x)|^p\,\text{d}x\Big)^{\frac{1}{p}},$$

for an arbitrary $p\ge1$. This simply means that $L^p[a,b]\subset L^1[a,b]$. Specifically, we have

$$L^2[a,b]\subset L^1[a,b],\qquad L^3[a,b]\subset L^1[a,b].$$

So, $1$ and $2$ are indeed equivalent. If we replace $f(x)$ with $f^2(x)$ and $g(x)=1$ in the inequality we obtain

$$\int_a^b |f(x)|^2\,\text{d}x\le \Big(\int_a^b |f(x)|^{2p}\,\text{d}x\Big)^{\frac{1}{p}},$$

and choosing $p=\frac{3}{2}$ gives

$$\int_a^b |f(x)|^2\,\text{d}x\le \Big(\int_a^b |f(x)|^{3}\,\text{d}x\Big)^{\frac{2}{3}}.$$

However, this last inequality implies that $L^3[a,b]\subset L^2[a,b]$ which is the converse of what the question says! Am I missing something here?

Furthermore, $f(x)=x^{-\frac{5}{12}}$ is $L^2[0,1]$ but not $L^3[0,1]$. So, I guess there should be a typo in the question.

Another example that I was thinking about was $f(x)=x^{-\frac{1}{2}}$. It seems that it is $L^3[1,\infty)$ but not $L^2[1,\infty)$. Converse of the previouse example! Can you shed some light on this?

Future readers can take a look at this post which is closely related.

Answer 1

Well, according to the comments, the inequality is called H$\ddot{\mathbf{\text{o}}}$lder inequality. The correct inclusion relation is

$$L^3[a,b]\subset L^2[a,b]\subset L^1[a,b].$$

More generally, if $p<q$ and $[a,b]$ is bounded then

$$L^q[a,b]\subset L^p[a,b].$$

For unbounded domains, this may not be true as the example at the end of question points out. This also implies that the H$\ddot{\text{o}}$lder inequality is not valid for such domains.