Find the fallacy in using the Cauchy–Schwarz inequality

Question

Let $\int_{a}^{b}\frac{f(x)}{x}dx=k$, wherein $f(x),a,b,k$ are positive. According to the Cauchy–Schwarz inequality:

$\int_{a}^{b}xf(x)dx=\int_{a}^{b}x^{2}\frac{f(x)}{x}dx\leq \left ( \int_{a}^{b}x^{4}dx\right )^{0.5} \left ( \int_{a}^{b}\left (\frac{f(x)}{x} \right )^{2}dx \right )^{0.5}$,

which implies

$\int_{a}^{b}xf(x)dx< \left ( \int_{a}^{b}x^{4}dx\right )^{0.5} \int_{a}^{b}\frac{f(x)}{x}dx= k\left ( \int_{a}^{b}x^{4}dx\right )^{0.5}=U$.

On the other hand:

$k=\int_{a}^{b}\frac{f(x)}{x}dx=\int_{a}^{b}\frac{1}{x^2}xf(x)dx\leq \left ( \int_{a}^{b}\frac{1}{x^4}dx\right )^{0.5} \left (\int_{a}^{b}\left( xf(x) \right)^2dx\right )^{0.5},$

which means that

$\int_{a}^{b}xf(x)dx>\left (\int_{a}^{b}\left( xf(x) \right)^2dx\right )^{0.5}\geq k\left ( \int_{a}^{b}\frac{1}{x^4}dx\right )^{-0.5}=L$

To sum up:
\begin{align} L=k\left ( \int_{a}^{b}\frac{1}{x^4}dx\right )^{-0.5}<\int_{a}^{b}xf(x)dx<k\left ( \int_{a}^{b}x^{4}dx\right )^{0.5}=U \end{align}

If $[a,b]=[1,2]$, $U=\sqrt{\frac{31}{5}}k>\sqrt{\frac{24}{7}}k=L$: OK

But if $[a,b]=[0.5,1]$, $U=\sqrt{\frac{31}{160}}k<\sqrt{\frac{3}{7}}k=L$, meaning that the upper-bound is smaller than the lower-bound!

In what step I am making a mistake?

Answer 1

This one is off: $$\int_{a}^{b}xf(x)dx< \left ( \int_{a}^{b}x^{4}dx\right )^{0.5} \int_{a}^{b}\frac{f(x)}{x}dx.$$ It should be (as you previously written, maybe it's just a copypaste error) $$\int_{a}^{b}xf(x)dx\leq \left ( \int_{a}^{b}x^{4}dx\right )^{0.5} \left ( \int_{a}^{b}\left (\frac{f(x)}{x} \right )^{2}dx \right )^{0.5}.$$

Now, there're some inequalities allowing to compare $L_1$ and $L_2$ norm on a compact, but it definitely involves some power of $(b-a)$.

Answer 2

It is not about Cauchy-Schwarz, this is not true: $ (\int_a^b(\frac{f(x)}{x})^2dx)^{0.5}<\int_a^b\frac{f(x)}{x}dx$. Take for instance, $f(x)=x$ and $0<(b-a)<1; a,b>0$.