Using Cauchy Schwarz to prove boundedness

Question

I'm trying to prove the boundedness on these improper integrals by using Cauchy-Schwarz's inequality.

$$ \int_{-\infty}^{c} \frac{h(t)}{t-x} \ dt, \int_{b}^{\infty} \frac{h(t)}{t-x} $$

Where $h(t) \in \mathtt{C}^1(\mathbb{R})$

I've done the following:

$$ \bigg{|} \int_{-\infty}^{c} \frac{h(t)}{t-x} \ dt \bigg{|}, \bigg{|} \int_{b}^{\infty} \frac{h(t)}{t-x}dt \bigg{|} < \int_{-\infty}^{\infty} \frac{|h(t)|}{|t-x|}dt < \bigg(\int_{-\infty}^{\infty} |h(t)|^2dt \bigg)^{1/2}\bigg(\int_{-\infty}^{\infty} |1/(t-x)|^2dt \bigg)^{1/2}$$

The first integral is bounded because of the boundedness of $h(t)$ since its a $C^1$ function. But the other one has to hold too, for the inequality to be true. And I don't know how to prove it.

EDIT: Regarding the comments, I rechecked the hypotheses of the exercise. I forgot there's this one:

$$\int_{-\infty}^{\infty} |h(t)|^2dt < \infty $$

Answer 1

There's no need to consider $\int_{-\infty}^\infty\frac{dt}{(t-x)^2}$. If $x<b$,$$\int_b^\infty\frac{dt}{(t-x)^2}=\left[\frac{1}{x-t}\right]_c^\infty=\frac{1}{b-x}.$$Similarly, if $x>c$,$$\int_{-\infty}^c\frac{dt}{(t-x)^2}=\left[\frac{1}{x-t}\right]_{-\infty}^c=\frac{1}{x-c}.$$Thus $\int_b^\infty\frac{h(t)dt}{t-x}$, $\int_{-\infty}^c\frac{h(t)dt}{t-x}$ are finite if $x$ satisfies suitable conditions and so does $h$. (Others have addressed what those conditions are.)

Answer 2

These improper integrals are not bounded in general for $h \in C^1(\mathbb{R})$. An easy example to see that is to take $h(t) = t - x$, which has indeed a continuous derivative. Then $$ \int_b^\infty \frac{h(t)}{t-x} \mathrm{d}t = \int_b^\infty \mathrm{d}t = \infty, $$ and the same works for the other integral. Nevertheless, your proof is good for $h \in L^2(\mathbb{R})$.