Find a non-zero function $f \in C(\mathbb{R}/\mathbb{Z} ; \mathbb{C})$ such that $||f||_2 = A$ and $||f||_\infty = B$.

fourier seriesfunctional-analysisreal-analysis

(Tao Vol.2, P.113, Q.5.2.3) If $f \in C(\mathbb{R}/\mathbb{Z} ; \mathbb{C})$ is a non-zero function, show that $0 < ||f||_2 < ||f||_\infty$. Conversely, if $0 < A \le B$ are real numbers, show that there exists a non-zero function $f \in C(\mathbb{R}/\mathbb{Z} ; \mathbb{C})$ such that $||f||_2 = A$ and $||f||_\infty = B$. (Hint: let $g$ be a non-constant non-negative real-valued function in $C(\mathbb{R}/\mathbb{Z} ; \mathbb{C})$, and consider functions $f$ of the form $f = (c + dg)^{1/2}$ for some constant real numbers $c,d >0$.)

$C(\mathbb{R}/\mathbb{Z} ; \mathbb{C})$ denotes a collection of continuous functions whose domain is the quotient space of $\mathbb{R}$ modulo $\mathbb{Z}$ and range is complex number $\mathbb{C}$.

$$||f||_2^2 = \int_{[0,1]}|f(x)|^2 \le \sup_{x \in [0,1)} |f(x)|^2 = ||f||_\infty^2. $$

I am struggling with the opposite direction. I cannot see how $f = (c+dg)^{1/2}$ help the proof. I would appreciate if you give some help.

Best Answer

You want: $$ \|f\|_2=A, \qquad\|f\|_{\infty}=B $$ Which means $$\|(c+dg)^{1/2}\|_2^2=A^2, \qquad\|(c+dg)^{1/2}\|_{\infty}^2=B^2$$ Try to express $\|(c+dg)^{1/2}\|_2^2$ and $\|(c+dg)^{1/2}\|_{\infty}^2$ in terms of $c,d$ and some suitable norms of $g$. Can you figure out how to conclude now? (Remember $|x|=x$ for $x\geq 0$)

EDIT:

We get the following linear system for $c,d$: $$\begin{cases} c+d\alpha=A^2\\ c+d\beta=B^2 \end{cases}$$ Where $$\alpha :=\int_{[0,1]}g = \|g\|_{1} \qquad \beta := \text{sup}_{[0,1]}g = \|g\|_{\infty}$$

  • CASE $A=B:$ Can you guess at a possible solution?

  • CASE $A\neq B$: Choosing $g$ so that $0<\alpha<\beta$ (is it possible?), the system admits one and only one solution: $$\begin{cases} c=(\beta A^2 - \alpha B^2)/(\beta-\alpha)\\ d=(B^2-A^2)/(\beta-\alpha) \end{cases}$$ We are almost done. We still have to be sure that $c,d$ be positive. The condition $d>0$ leads to $A<B$, which is automatic in this case. The condition $c>0$ leads to: $$ \frac{\alpha}{\beta}<\frac{A^2}{B^2} $$ Since $A,B$ are given, we have to look for suitable $g$ so that the inequality above is satisfied (remember that $\alpha, \beta$ depend on $g$). This can be achieved in at least two different ways: looking for an explicit $g$ satisfying this condition (this is a good exercise) or using the fact that the norms $\| \cdot \|_{\infty}$ and $ \| \cdot \|_{1}$ are not equivalent.

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