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.