0
$\begingroup$

Using the following Cauchy-Schwarz Inequality for integrals,

$$\int_c^d\int_a^bf^2(x,y)dxdy\int_c^d\int_a^bg^2(x,y)dxdy \geq \Bigg[\int_c^d\int_a^bf(x,y)g(x,y)dxdy\Bigg]^2$$

Prove: $$\sqrt{\int_c^d\int_a^b[f(x,y) + g(x,y)]^2dxdy} \leq \sqrt{\int_c^d\int_a^bf^2(x,y)dxdy} + \sqrt{\int_c^d\int_a^bg^2(x,y)dxdy}$$

I am struggling with this proof. I tried starting by expanding the left handside and constructing an inequality by a substitution of the C-S inequality but this did not seem to help.

$\endgroup$

1 Answer 1

0
$\begingroup$

Putting double integrals seems superfluous so lets use just one. We use this form of Cauchy-Schartz (which is just Hölder for $p=q=2$): \begin{align*} \left(\int fg\right)^2 \leq \int f^2 \int g^2. \end{align*}

It is equivalent to prove the square of the equality: \begin{align*} \int (f+g)^2 \leq \int f^2 + \int g^2 + 2\sqrt{\int f^2} \sqrt{\int g^2}. \end{align*}

Note that we may expand the LHS and apply Cauchy-Schwartz on the cross-term: \begin{align*} \int (f+g)^2 = \int f^2 + \int g^2 + 2\int fg \leq \int f^2 + \int g^2 + 2\sqrt{\int f^2}\sqrt{\int g^2}. \end{align*}

Taking square roots gives the desired result.

$\endgroup$
1
  • $\begingroup$ Thankyou. I sort of did this in reverse. I arrived at your final step but failed to realise that I had proved the square of the inequality! $\endgroup$ May 5, 2020 at 1:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .