$f(x)$ continuous in $[0,\infty)$. Both $\int^\infty_0f^4(x)dx, \int^\infty_0|f(x)|dx$ converge. Prove that $\int^\infty_0f^2(x)dx$ converges

Question

$f(x)$ continuous in $[0,\infty)$. Both $\int^\infty_0f^4(x)dx, \int^\infty_0|f(x)|dx$ converge.

I need to prove that $\int^\infty_0f^2(x)dx$ converges.

Knowing that $\int^\infty_0|f(x)|dx$ converges, I can tell that $\int^\infty_0f(x)dx$ converges.

Other than that, I don't know anything. I tried to maybe integrate by parts but I got to nowhere

$\int^\infty_0f^4(x)dx = \int^\infty_0f^2(x)|f(x)|\cdot |f(x)| dx$..

Answer 1

By AM/GM $$3f(x)^2\le f(x)^4+|f(x)|+|f(x)|.$$

Answer 2

Hint Notice that if $a\ge1$ then $a^4\ge a^2$ and if $ \vert a\vert\le1$ then $a^2\le \vert a\vert$ so in all cases $$a^2\le \max(\vert a\vert,a^4)$$

Answer 3

Here is a proof. Let $$ A \triangleq \{x\in[0,\infty) : |f(x)|\leq 1\}, \ \text{and} \ A^c \triangleq \{x\in[0,\infty) : |f(x)|>1\}. $$

Write, $$ \int_0^{\infty}f^2(x) dx = \underbrace{\int_Af^2(x) dx}_{\triangleq M} + \underbrace{\int_{A^c}f^2 (x) dx}_{\triangleq N}. $$ Note that $M$ is bounded, since $$ \int_A f^2(x) dx \leq \int_A |f(x)| dx \leq \int_{[0,\infty)}|f(x)|dx<\infty. $$ For $N$, we'll use Cauchy-Schwarz inequality: \begin{align} \left(\int_{[0,\infty)}f^{4}(x)dx\right)\left(\int_{[0,\infty)}|f(x)|dx\right) & \geq \left(\int_{[0,\infty)}|f(x)|^{\frac{5}{2}}dx\right)^2 \\ & \geq \left(\int_{A^c}|f(x)|^{\frac{5}{2}}dx\right)^2 \\ &\geq \left(\int_{A^c}|f(x)|^2 dx\right)^2 \end{align} hence, $N$ is also bounded. Therefore, we are done.

Edit

To address Open Ball's comment below, let $f \in L^p(X) \cap L^q(X)$ on some measure space $(X,\mathcal{F},\mu)$. Then note that, $$ |f|^{\lambda p + (1-\lambda)q} \leq \lambda|f|^p + (1-\lambda)|f|^q, $$ using the convexity of the map $x \mapsto a^x$ ($a>0$). With this, one can obtain weight $\lambda$ to express any $r \in [p,q]$ as $\lambda p + (1-\lambda)q$. Taking the integral of the inequality above with respect to underlying measure $\mu$, and keeping in mind the monotonicity of integral, we get $$ \int |f|^r d\mu \leq \lambda \int |f|^p d\mu + (1-\lambda) \int |f|^q d\mu \implies ||f||_{L^r(X)}^r \leq \lambda ||f||_{L^p(X)}^p + (1-\lambda)||f||_{L^q(X)}^q. $$

Hence, if $f \in L^p(X) \cap L^q(X)$, then for every $p\leq r \leq q$, $f \in L^r(X)$.

Answer 4

Disclaimer: this is an overkill.

Here's a general fact: if $(X, \Sigma, \mu)$ is a measure space, we have for $1\le p<q \le \infty$, $L^p(X) \cap L^q(X) \subset L^r(X)$ for all $p \le r \le q$.

Here, $f \in L^1([0,\infty)) \cap L^4([0,\infty))$, so it is in $L^r([0,\infty))$ for each $r \in [1,4]$.

Answer 5

Improved version of Lord Shark's answer: for any $K>0$ we have, by AM-GM, $$ 3 f(x)^2 \leq K^2 f(x)^4 + \frac{1}{K}|f(x)|+ \frac{1}{K}|f(x)| \tag{1}$$ from which it follows that $$ \int_{0}^{+\infty}f(x)^2\,dx \leq \frac{K^2}{3}\int_{0}^{+\infty} f(x)^4\,dx + \frac{2}{3K}\int_{0}^{+\infty}|f(x)|\,dx \tag{2} $$ and the RHS of $(2)$ is minimized by choosing $K=\left(\frac{\int_{0}^{+\infty}|f(x)| dx}{\int_{0}^{+\infty}f(x)^4 dx}\right)^{1/3}$, leading to the following form of Holder's inequality: $$ \int_{0}^{+\infty}f(x)^2\,dx \leq \left(\int_{0}^{+\infty}f(x)^4\,dx\right)^{1/3}\left(\int_{0}^{+\infty}|f(x)|\,dx\right)^{2/3}.\tag{3}$$ The same path can be followed for proving the more general Riesz-Thorin theorem, essentially equivalent to the log-convexity of the map $p\mapsto \|f\|_p=\left(\int_{0}^{+\infty}|f(x)|^p\,dx\right)^{1/p}$.

Answer 6

Let $A = \{|f|\le 1\}, B = \{|f|>1 \}.$ Then $\int_A f^2 \le \int_A |f|<\infty$ and $\int_B f^2 \le \int_B f^4<\infty.$ The result follows.