Is L2 norm always smaller than L1 norm even in infinite dimensional space?

Question

I know that in infinite dimensional space, the inequality for L norms in discrete cases don't always work. For example, L-infinity norm can be greater than any other L-norm in special cases.

Now, I wonder if L2 norm is always smaller than L1 norm in infinite dimensional space? When p is less than infinity, does the general inequality ‖∗‖q≤‖∗‖p whenever p≤q holds?

In a geometrical sense, L2 norm should be always smaller than L1 norm though. By the way, Lq space $\in$ Lp space whenever q≤p right?

Answer 1

No. For instance, $f(x):=\frac{1}{\sqrt{x}}\in L^1(0,1)\setminus L^2(0,1)$, hence $\|f\|_{L^2}=\infty \not \leq \|f\|_{L^1}<\infty$.

In rough terms, when increasing the value of $p$, the $\|\cdot\|_{L^p}$ norm of function $f$ will be less sensitive to the rate of decrease of $f$ at infinity, but more sensitive to its rate of increase near any singularities. Therefore the inequality $L^p\subset L^q$ if $p\leq q$ is only true when functions cannot have singularities, such as in the case of $\ell^p=L^p(\mathbb{N})$. Conversely, when the domain of the functions has finite measure (and hence there are no issues with decrease at infinity), the inclusions are reversed, i.e. $L^q\subset L^p$ if $p\leq q$.

Answer 2

The situation with $L^p$ in general is much less straightforward than you might hope. First, the assertion that the 2-norm is always less than the 1-norm of a function is false. To see this, consider $X = [0, \frac{1}{2}]$, with Lebesgue measure $m$. Let $f: X \rightarrow \mathbb{C}$ be the function defined such that $f(x) = 1$ for all $x \in X$. Then $$||f||_1 = \int_{X} |f(x)| dx = \frac{1}{2}.$$ On the other hand, $$||f||_2 = \bigg(\int_{X}|f(x)|^2 dx\bigg)^{\frac{1}{2}} = \sqrt{\frac{1}{2}} > \frac{1}{2}. $$ So it is evidently false that the 2-norm of a function is always less than its 1-norm.

It is also false that $L^q \subset L^p$ whenever $p \leq q$, even when $p \geq 1$. To see this, just consider $f: (1,\infty) \rightarrow \mathbb{R}$ defined by $f(x) = \frac{1}{x}$. This function is in $L^2((1,\infty))$ but not $L^1((1,\infty))$.

Answer 3

Take $f_t = {1 \over t} 1_{[0,t]}$ on $[0,2]$. Then $\|f_t\|_1 = 1$ for all $t>0$, but $\|f_t\|_2 = {1 \over \sqrt{t}}$.