Holder’s Inequality for integrals (non-negative functions)

definite integralsholder-inequalityintegral-inequalityreal-analysissolution-verification

Let's recall Young's Inequality.

Statement: Let $u, v \geqslant 0$, and $p, q \in (0, \infty)$ such that
$$ uv \leqslant \frac{u^p}{p} + \frac{v^q}{q}$$

$\blacksquare~$Problem: Let $p,q$ (Holder Conjgates) be positive real numbers satisfying
$$\frac{1}{p} + \frac{1}{q} =1 $$
Then prove the following

  • If $f,g$ are Riemann integrable non-negative functions, then
    $$
    \int_a^b fg~ \mathrm{d}x \leqslant \left\{\int_a^b f^p ~\mathrm{d}x\right\}^{\frac{1}{p}} \left\{\int_a^b g^q ~\mathrm{d}x\right
    \}^{\frac{1}{q}}
    $$

$\blacksquare~$Solution: The problem is trivial (equality holds) when the value of both integrals is $0$. Then let's consider the first case (reduced) as

$\bullet~$ Case $1$: If $$ \int_a^b f^p ~\mathrm{d}x = \int_a^b g^q ~\mathrm{d}x = 1 \implies \left\{\int_a^b f^p ~\mathrm{d}x \right\}^\frac{1}{p} \left\{ \int_a^b g^q ~\mathrm{d}x \right\}^\frac{1}{q} = 1 $$
then from Young's Inequality we have that
\begin{equation*}
\begin{split}
fg &\leqslant \frac{f^p}{p} + \frac{g^q}{q}\\
\implies \int_a^b fg ~\mathrm{d}x &\leqslant \frac{1}{p} \int_a^b f^p~\mathrm{d}x + \frac{1}{q} \int_a^b g^q~\mathrm{d}x\\
\implies \int_a^b fg~\mathrm{d}x &\leqslant \frac{1}{p} + \frac{1}{q} = 1 = \left\{\int_a^b f^p ~\mathrm{d}x \right\}^\frac{1}{p} \left\{ \int_a^b g^q ~\mathrm{d}x \right\}^\frac{1}{q}
\end{split}
\end{equation*}
Hence, we have established the inequality for $\textbf{Case 1.}$

$\bullet~$ Case $2$:
The general case, i.e.,
$$ \int_a^b f^p \mathrm{d}x ~\text{ and }~ \int_a^b g^q \mathrm{d}x \neq 1$$
Then let's assume that
\begin{align}
\label{equation 4}
\int_a^b f^p \mathrm{d}x = \alpha^p ~\text{ and }~ \int_a^b g^q \mathrm{d}x = \beta^q \quad \text{for }~\alpha, \beta \in \mathbb{R}^{+}
\end{align}
Thus we can easily see that
\begin{align}
\label{general case}
\int_a^b \left(\frac{f}{\alpha}\right)^p \mathrm{d}x = 1 ~\text{ and }~ \int_a^b \left(\frac{g}{\beta}\right)^q \mathrm{d}x = 1 \quad \text{for }~\alpha, \beta \in \mathbb{R}^{+}
\end{align}
Then from Young's Inequality we have that
\begin{equation*}
\begin{split}
\left( \frac{fg}{\alpha \beta} \right) &\leqslant \left( \frac{f^p}{p \cdot \alpha^p} \right) + \left( \frac{g^q}{q \cdot \beta^q} \right)\\
\implies \int_a^b \left( \frac{fg}{\alpha \beta} \right) ~\mathrm{d}x &\leqslant \frac{1}{p} \int_a^b \left(\frac{f}{\alpha}\right)^p~\mathrm{d}x + \frac{1}{q} \int_a^b \left( \frac{g}{\beta} \right)^q~\mathrm{d}x \\
\implies \frac{1}{\alpha \beta} \int_a^b fg~\mathrm{d}x &\leqslant \frac{1}{p} + \frac{1}{q} = 1 \quad \\
\implies \int_a^b fg ~\mathrm{d}x &\leqslant \alpha \beta = (\alpha^p)^{\frac{1}{p}} \cdot (\beta^q)^\frac{1}{q} \\
\implies \int_a^b fg ~\mathrm{d}x &\leqslant \left\{\int_a^b f^p ~\mathrm{d}x \right\}^\frac{1}{p} \left\{ \int_a^b g^q ~\mathrm{d}x \right\}^\frac{1}{q} \quad [\text{From our construction}]
\end{split}
\end{equation*}
Thus, we have proved $\textbf{Case 2.}$ The equality holds when $\beta^q \cdot f^p = \alpha^p \cdot g^q$.

Hence we are done!

Please check for glitches and if it's fine, it'll be great if I get another way of solution.

Thanks in Advance Guys! 🙂

Best Answer

It is not clear that your proof covers all possible cases. Also it can be simplified a bit by using previous results instead of repeating arguments.

You say

The problem is trivial (equality holds) when the value of both integrals is $0$.

but what you actually need is that equality holds if either of the integrals on the right-hand side is zero. That is still correct because it implies that either $f$ or $g$ is zero almost everywhere on $[a,b]$, so that the integral on the left-hand side is zero as well.

Now you can consider the case that both $\int_a^b f(x) \, dx$ and $\int_a^b g(x) \,dx$ are non-zero. You do that by considering two cases:

Case 1: $\int_a^b f(x) \, dx = \int_a^b g(x) \,dx = 1$.

Case 2: The general case, i.e., $\int_a^b f(x) \, dx$ and $\int_a^b g(x) \,dx \ne 1$.

If the second case means that both integrals are not equal to one, then your proof does not cover the cases that one integral is equal to one and the other is not.

Actually your proof of the “Case 2” works as long as both integrals on the right are different from zero, which means that “Case 1” is not needed.

Note also that you use Young's inequality in both cases, so you are repeating some arguments.

What you can do instead (to clarify and simplify) your proof, is one of the following.

Either:

  1. Prove Hölder's inequality for the case that $\int_a^b f(x) \, dx = 0 $ or $\int_a^b g(x) \, dx = 0$.
  2. Then prove Hölder's inequality for the case that $\int_a^b f(x) \, dx \ne 0 $ and $\int_a^b g(x) \, dx \ne 0$. This would be what you wrote in your “Case 2,” using Young's inequality.

Or:

  1. Prove Hölder's inequality for the case that $\int_a^b f(x) \, dx = 0 $ or $\int_a^b g(x) \, dx = 0$.
  2. Then prove Hölder's inequality for the case that $\int_a^b f(x) \, dx = 1 $ and $\int_a^b g(x) \, dx = 1$. This would be what you wrote in your “Case 1,” using Young's inequality.
  3. Finally prove Hölder's inequality for the case that $\int_a^b f(x) \, dx \ne 0 $ and $\int_a^b g(x) \, dx \ne 0$. Here you should not repeat the argument from part 2, but use that result instead.

Let me clarify what I mean with “using that result instead.“ If $\alpha > 0$ and $\beta > 0$ are defined such that $$ \int_a^b f(x)^p \, dx = \alpha^p \, , \, \int_a^b g(x)^q \, dx = \beta^q $$ then (as you say) $$ \int_a^b \left(\frac{f(x)}{\alpha}\right)^p \, dx = \int_a^b \left(\frac{g(x)}{\beta}\right)^q \, dx = 1 \, . $$ From the previous part we know that Hölder's inequality holds for the functions $f/\alpha$ and $g/\beta$, i.e. that $$ \int_a^b \frac{f(x)g(x)}{\alpha \beta} \, dx \le 1 \, . $$ It follows that $$ \int_a^b f(x) g(x) \, dx \le \alpha \beta $$ and that is exactly Hölder's inequality for $f$ and $g$.

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