Proof of Holder's Inequality in Multivariable Calculus

Question

I am self studying Calculus 3 (Multivariate/Vector Calculus) as a hobby and I came across this question. I graduated with a degree in chemistry, so I know a bit of math, but not too much to answer this question. Haha.

Anyway, I'm trying to prove what seems to be called the Holder's Inequality in Marsden and Tromba's Vector Calculus.

$$\left\lvert \displaystyle\int_0^1 f(x)g(x) \, dx \right\rvert \leq \sqrt{\displaystyle\int_0^1 [f(x)]^2 \, dx}\sqrt{\displaystyle\int_0^1 [g(x) ]^2\, dx}$$

My initial guess was to use the inequality between arithmetic mean and geometric mean, and I have tried to get rid of the factor 2 that appears after using the relationships between those two means, but to no avail.

$$ \sqrt{f(x)g(x)} \leq \dfrac{1}{2}({f(x)+g(x)}) $$

The question specifically states that:

1. We should think about functions as vectors, satisfying scaling behavior and closed under addition, and
2. We should introduce the idea of the inner product to functions, namely:

$$ f \cdot g =\displaystyle\int_0^1 f(x)g(x) \, dx $$ and

3. We should verify that the integral as described in 2 satisfies properties of inner products:

$$ \text{a)} (\alpha\mathbf{x}+\beta\mathbf{y})\cdot \mathbf{z} = \alpha\mathbf{x}\cdot\mathbf{z}+\beta\mathbf{y}\cdot\mathbf{z}\\ \text{b)} \mathbf{x}\cdot\mathbf{y}=\mathbf{y}\cdot\mathbf{x}\\ \text{c)} \mathbf{x}\cdot\mathbf{x}\ge0\\ \text{d)} \mathbf{x}\cdot\mathbf{x}=0 \iff \mathbf{x}= \mathbf{0}\\ $$

So I was thinking of asking for a proof for the first inequality, preferably at a level of someone who knows ordinary differential equations and linear algebra, but no real analysis? Thank you so much in advance.

Answer 1

In the context of linear algebra we have the Cauchy-Schwarz's inequality:

Theorem: Let $V$ be an inner product space. If $v,w\in V$, then $|\langle v,w\rangle|^{2}\leqslant \langle v,v\rangle \langle w,w\rangle$, where $\langle \cdot,\cdot\rangle$ denote the inner product.

In order to prove the theorem, you can prove the theorem directly using only linear algebra. Then, you can establish (prove that the defined product is an inner product) in the vector space $V=C([0,1];{\bf R})$ on which we have defined the inner product $\langle f,g\rangle=\int_{0}^{1}f(x)g(x)\, dx$. Then, you can conclude for the theorem $$\left|\int_{0}^{1}f(x)g(x)\, dx\right|\leqslant \sqrt{\int_{0}^{1}f(x)f(x)\,dx}\sqrt{\int_{0}^{1}g(x)g(x)\, dx},\quad (\ast)$$

Proposition: In the vector space $V=C([0,1]; {\bf R})$ the map $\langle f,g\rangle =\int_{0}^{1}f(x)g(x)\, dx$ define an inner product.
Proof.

• $\langle f+g,z\rangle=\int_{0}^{1}(f+g)z=\int_{0}^{1}fg+\int_{0}^{1}gz=\langle f,g\rangle+\langle g,z\rangle$, for all $f,g,z\in V$.
• $\langle af,g\rangle=\int_{0}^{1}(af)g=a\int_{0}^{1}fg=a\langle f,g\rangle$ for all real number $a$ anf for all $f,g\in V$.
• $\langle f,g\rangle=\int_{0}^{1}fg=\int_{0}^{1}gf=\langle g,f\rangle$, for all $f,g\in V$.
• If $f\not=0$, then $f^{2}$ is bounded on the closed-bounded set $[0,1]$ by continuity and then $\langle f,f\rangle=\int_{0}^{1}f^{2}>0$.

Therefore, $\langle \cdot,\cdot\rangle$ is a inner product in $V$ and then $(\ast)$ is follows of the theorem.

Answer 2

You can actually recover a proof of Cauchy-Schwarz (and more generally Holder's inequality) using Young's inequality. For Cauchy-Schwarz specifically, this is the very simple inequality (which is equivalent to) AGM: \begin{align} ab\leq \frac{a^2+b^2}{2}\tag{$*$}, \end{align} for all $a,b\in\Bbb{R}$. This follows of course by noting that $0\leq (a-b)^2=a^2-2ab+b^2$, and rearranging. Now, we have for any integrable functions $f,g:I\to\Bbb{R}$ (where $I=[\alpha,\beta]$ is an interval for instance), \begin{align} \int_If(x)g(x)\,dx\leq\int_I\frac{f(x)^2+g(x)^2}{2}\,dx=\frac{1}{2}\int_If(x)^2\,dx+\frac{1}{2}\int_Ig(x)^2\,dx.\tag{$**$} \end{align} And now, we can actually exploit a symmetry-imbalance in this inequality. The above inequality is true for all pairs of (integrable) functions $(f,g)$. So, for any $\lambda>0$, let us apply the inequality above to the pair $\left(\lambda f,\frac{g}{\lambda}\right)$. This gives \begin{align} \int_I\left(\lambda f(x)\right)\cdot\left(\frac{g(x)}{\lambda}\right)\,dx&\leq \frac{1}{2}\int_I\left(\lambda f(x)\right)^2\,dx+\frac{1}{2}\int_I\left(\frac{g(x)}{\lambda}\right)^2\,dx, \end{align} which says \begin{align} \int_If(x)g(x)&\leq \frac{\lambda^2}{2}\int_If(x)^2\,dx+\frac{1}{2\lambda^2}\int_Ig(x)^2\,dx\tag{$***$} \end{align} The inequality $(***)$ is thus true for all functions $(f,g)$ and all $\lambda>0$. This is what I meant by symmetry imbalance above: the LHS of the $(**)$ didn't change when I made the transformation $(f,g)\to \left(\lambda f,\frac{g}{\lambda}\right)$, but the RHS of $(**)$ did change. This symmetry imbalance can be exploited to prove a stronger inequality, by choosing a value of $\lambda$ which minimizes the expression on the right of $(***)$.

Let's do this now: if either of $\int_If(x)^2\,dx$ or $\int_Ig(x)^2\,dx$ are equal to $0$, then Cauchy-Schwarz is obvious, so let us consider the case where these integrals are strictly positive. Then, think of the RHS of $(***)$ as a function of $\lambda\in (0,\infty)$. If $\lambda\to 0^+$ or $\lambda\to \infty$, then the RHS blows up, which proves there must be a strict global minimum point at some $\lambda_0\in (0,\infty)$. Now, it is a simple calculus exercise for you to find the derivative with respect to $\lambda$, set that equal to $0$, and deduce that $\lambda_0^2=\sqrt{\frac{\int_Ig^2}{\int_If^2}}$, so plugging in this value of $\lambda_0^2$ into $(***)$ gives \begin{align} \int_If(x)g(x)\,dx&\leq \sqrt{\int_If(x)^2\,dx\int_Ig(x)^2\,dx}. \end{align} Ok, this is not quite Cauchy-Schwarz, but you can recover it by applying the above inequality to $|f|$ and $|g|$ (since we know the above inequality holds for all pairs of functions) \begin{align} \left|\int_If(x)g(x)\,dx\right|&\leq\int_I|f(x)g(x)|\,dx\leq\sqrt{\int_I|f(x)|^2\,dx\int_I|g(x)|^2\,dx}. \end{align} (The first inequality is obvious because $-|f(x)g(x)|\leq f(x)g(x)\leq |f(x)g(x)|$, so it follows by monotonicity of integrals).

So I'm guessing that when you tried to use AGM you didn't continue further because the RHS of AGM is the arithmetic mean (a sum) whereas the RHS of Cauchy-Schwarz is actually a product, and so you weren't able to continue. Well, as shown above, AGM is a "wasteful" inequality in the sense that it's not the best possible (it introduces a scaling imbalance on the two sides of the inequality).

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You can actually prove Holder's inequality (which is a vast generalization of Cauchy-Schwarz, and one of the most important "elementary" inequalities in math I'd say) along the exact same lines (except you have to handle a few more exceptional cases). The only real extra ingredient you need is Young's inequality: \begin{align} ab&\leq \frac{a^p}{p}+\frac{b^q}{q}, \end{align} for all $a,b\in [0,\infty)$, and all $1<p<\infty$, with $q$ being the Holder conjugate. Then, you get \begin{align} \left|\int_If(x)g(x)\,dx\right|&\leq \int_I|f(x)|\cdot |g(x)|\,dx\leq\int_I\frac{|f(x)|^p}{p}\,dx+\int_I\frac{|g(x)|^q}{q}\,dx. \end{align} Now, using the above inequality with $\lambda f$ and $\frac{g}{\lambda}$, we get \begin{align} \int_I|f(x)g(x)|\,dx&\leq\frac{\lambda^p}{p}\int_I|f(x)|^p\,dx+\frac{1}{q\lambda^q}\int_I|g(x)|^q\,dx. \end{align} Barring some exceptional cases, you can differentiate with respect to $\lambda$, find the minimum point $\lambda_0$, plug it back in, and use the Holder-conjugate condition $\frac{1}{p}+\frac{1}{q}=1$ to obtain Holder's inequality: \begin{align} \int_I|f(x)g(x)|\,dx&\leq\left(\int_I|f(x)|^p\,dx\right)^{1/p}\left(\int_I|g(x)|^q\,dx\right)^{1/q}, \end{align} i.e $\|fg\|_{L^1(I)}\leq \|f\|_{L^p(I)}\|g\|_{L^q(I)}$.

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Terry Tao has a very interesting blog post which you may find an interesting read (atleast the first part should be understandable at a calculus level): Amplification, arbitrage, and the tensor power trick.