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I'm learning about Fourier analysis and need help with the following problem (which is part of a subchapter on $L^p$ spaces):

Using the Cauchy-Schwarz inequality show that if $f \in L^2[-\pi, \pi]$, then $(1)$ $f \in L^1[-\pi, \pi]$ and $(2)$ $\|f\|_{L^1} \leq \sqrt{2\pi} \|f\|_{L^2}$.

My work and thoughts:

$(1)$ We note that if $f \in L^2[-\pi, \pi]$, then the inequality $\|f\|_{L^1} \leq \sqrt{2\pi} \|f\|_{L^2}$ implies $\|f\|_{L^1} < \infty$ and hence $f \in L^1[-\pi, \pi]$.

What is bothering me about my proof for $(1)$ is that I'm using $(2)$ which I have not yet prove. Is my work correct and/or is there another way of showing $(1)$ without making use of $(2)$?

$(2)$ This is the part where I should apply the Cauchy-Schwarz inequality using the well-known "multiplication by one trick": $\int |f|^2 = \int g h$ where $g = |f|^2$ and $h = 1$. Since it is the first time I'm using the C-S inequality in the context of $L^p$ norms I'm having some difficulties applying it. Any help on this problem would be greatly appreciated.

glpsx
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2 Answers2

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The multiplication by 1 trick in action :

$$\int_{-\pi}^\pi |f(x)| dx = \left|\int_{-\pi}^\pi 1|f(x)| dx \right| $$

Then you apply Cauchy-Schwartz

$$ \leq \sqrt{\int_{-\pi}^\pi 1^2 dx } \sqrt{\int_{-\pi}^\pi |f(x)|^2 dx } $$

The first integral is easy to calculate, and the second is the $L^2$ norm

$$\leq \sqrt{2\pi} \|f\|_{L^2}$$

And this inequality is better than the one in your question , so I suppose a typo in your question

Tryss
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  • Great! Thank you for your detailed and clear explanation. It is extremely helpful and understandable. – glpsx Apr 05 '16 at 23:34
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In general, suppose that $X$ is finite measure space, and $p<q$. If $f\in L^q(X,\mu)$ then $f\in L^p(X,\mu)$, in fact $$|f|_p\leqslant \mu(X)^{1/p-1/q} |f|_q$$

Note that the first statement follows from the inequality stated, in fact this is how we prove such claim.

This can be proven directly with Hölder's inequality, much like Tryss's answer uses the particular case of Schwarz inequality. Thus taking $p=1,q=2$, $\mu$ the Lebesgue measure and $X=[-\pi,\pi]$ we have $\mu(X)=2\pi$, and the inequality reads

$$|f|_1\leqslant (2\pi ) ^{1/2} |f|_2$$

Pedro
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