1
$\begingroup$

I have proven something that is definitely not true (Lemma 2), which is why I am intersted where I err.

Definition Let $C(\mathbb{R}/\mathbb{Z},\mathbb{C})$ be the set of all continuous $\mathbb{Z}$-periodic functions from $\mathbb{R}$ to $\mathbb{C}$. We define inner product by $$\langle f,g\rangle = \int_{[0,1]}f\bar{g}$$ and $L^2$ norm by $$\|f\|_2 = \sqrt{\langle f,f \rangle}$$

Lemma 1. I have already proven all the properties of inner product, such as (a) positivity (b) hermitian property (c) linearity in the first variable and antilinearity in the second.

Lemma 2. I have to demonstrate the Cauchy-Schwarz inequality $$|\langle f,g \rangle | \leq \|f\|_2 \cdot \|g\|_2$$ But somehow magically I managed to show that they are always equal

$\begin{array}{ll} \iff \sqrt{\langle f,g \rangle \bar{\langle f,g \rangle }} = \sqrt{\langle f,f \rangle \langle g,g \rangle } \quad \text{by definition}\\ \iff \sqrt{\langle f,g \rangle \langle g,f \rangle } = \sqrt{\langle f,f \rangle \langle g,g \rangle } \quad \text{hermitian property}\\ \iff \sqrt{ \langle \langle f,g \rangle g,f \rangle } = \sqrt{\langle f\langle g,g \rangle,f \rangle } \quad \text{linearity}\\ \iff \sqrt{ \langle \langle fg,g \rangle,f \rangle } = \sqrt{\langle \langle fg,g \rangle,f \rangle } \quad \text{linearity}\\ \end{array}$

Where does my mistake dwell? Also, any help on how to proceed would be appreciated.

$\endgroup$
1
$\begingroup$

Your mistake is the last step. Linearity allows a scalar $k \in \Bbb{F}$ enters into the bracket $\langle u,v \rangle$ $$k\langle u,v \rangle = \langle ku,v \rangle,$$ but not a vector $w \in V$ $$w\langle u,v \rangle = \langle wu,v \rangle. \tag{wrong!}$$

Here, the $V = C(\mathbb{R}/\mathbb{Z},\mathbb{C})$, $\Bbb{F} = \Bbb{C}$, $u = f$ and $v = g$.

$\endgroup$
  • $\begingroup$ I see now! Would you give me a hint on how to proceed then? :) $\endgroup$ – Cebiş Mellim Feb 26 at 8:57
  • 1
    $\begingroup$ The proof is classic: consider $||u - kv||^2 \ge 0$ and expand it as a quadratic polynomial in $k$. The discriminant, which contains the terms in Cauchy-Schwarz inequality, should be nonpositive. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Feb 26 at 9:19
1
$\begingroup$

Linearity only allows you to pull out constants from inner products. It appears the in the last implication you have used linearity functions in stead of constants. In fact, you cannot even talk about inner product of $fg$ with $g$ because $fg$ need not be in $L^{2}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.