On a finite measure space, can we bound $\lVert f\rVert_1$ given $\lVert f\rVert_2$?

Question

Suppose $f$ is a positive measurable function on a finite measure space. If we know $\lVert f\rVert_2\leq\epsilon$, can we bound $\lVert f\rVert_1$ above?

In the space of real sequences with the counting measure this fails with a counterexample based on the harmonic series. For finite real vectors the result is true. I can't quite get the harmonic-series based counterexample to work in a probability space such as $\mathbb R$ equipped with a normal distribution, and I'm wondering if in fact the measure being finite is enough to give us the result above.

Answer 1

$$\int|f|=\int|f|\cdot1\leq\bigg(\int|f|^2\bigg)^{1/2}\bigg(\int 1^2\bigg)^{1/2}=||f||_2\cdot \mu(X)^{1/2}.$$

More generally, for any $0<p<q<\infty$ we have, $$f\in L^q\implies\int|f|^p=\int|f|^p\cdot1\leq \big|\big||f|^p\big|\big|_{q/p}\cdot ||1||_{q/(q-p)}=||f||^p_q\cdot \mu(X)^{(q-p)/q} $$$$\text{ That is }||f||_p\leq ||f||_q\cdot \mu(X)^{(1/p)-(1/q)}$$$$\text{ using Holder's inequality with exponents }q/p\text{ and }q/(q-p).$$

In case $q=\infty$ we have $$||f||^p_p=\int|f|^p\leq ||f||^p_\infty\int1=||f||^p_\infty\cdot \mu(X).$$

So, if $\mu(X)<\infty$ we have $||f||_p\leq ||f||_q\cdot \mu(X)^{\frac{1}{p}-\frac{1}{q}}$ for any $0<p<q\leq \infty$

Answer 2

Yes you can. That can be achieve by direct application of Holder's inequality.

For the specific case at hand, if $f\in L_2(\mu)$ $$\|f\|_1=\int_X|f|\,d\mu=\int_X|f|\mathbb{1}\,d\mu\leq \|f\|_2\|\mathbb{1}\|_2=\|f\|_2(\mu(X))^{1/2}\leq(\mu(X))^{1/2}\varepsilon$$

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More generally, if $0<s<r$ and $f\in L_r(\mu)$, then $|f|^s\in L_{r/s}(\mu)$ and so $$ \begin{align} \|f\|^s_s=\int_X|f|^s\,d\mu&=\int_X|f|^s\mathbb{1}\,d\mu\leq\||f|^s\|_{r/s}\|\mathbb{1}\|_{r(r-s)}\\ &=\|f\|^s_r\,(\mu(X))^{1-\frac{s}{r}} \end{align} $$ whence one gets that $$ \|f\|_s\leq\|f\|_r\,(\mu(X))^{\frac1s-\frac1r} $$