Reading some article a while ago I read the following: (here $H^2$ represents the Hardy space)

Let $f\in H^2$ be such that $f(0)=1$, and let $0<\lvert\lambda\rvert<1$, then $$\lVert f(\lambda z)\cdots f(\lambda^{n-1}z)\rVert _\infty \lVert f\rVert _2\leq \prod_{n\geq 1} \lVert f(\lambda^{n-1}z)\rVert_2$$

for every $n\geq 1$.

This seems a bit 'weird' to me, as usually the $2$-norm is bounded above by the infinity norm, not the other way as here.

I would greatly appreciate any help.

Reading some article a while ago I read the following: (here $H^2$ represents the Hardy space)

Let $f\in H^2$ be such that $f(0)=1$, and let $0<\lvert\lambda\rvert<1$, then $$\lVert f(\lambda z)\cdots f(\lambda^{n-1}z)\rVert _\infty \lVert f\rVert _2\leq \prod_{m\geq 1} \lVert f(\lambda^{m-1}z)\rVert_2$$

for every $n\geq 1$.

This seems a bit 'weird' to me, as usually the $2$-norm is bounded above by the infinity norm, not the other way as here.

I would greatly appreciate any help.

Reading some article a while ago I read the following: (here $H^2$ represents the Hardy space)

Let $f\in H^2$ be such that $f(0)=1$, and let $\lvert\lambda\rvert<1$, then $$\lVert f(\lambda z)\cdots f(\lambda^{n-1}z)\rVert _\infty \lVert f\rVert _2\leq \prod_{n\geq 1} \lVert f(\lambda^{n-1}z)\rVert_2$$

for every $n\geq 1$.

This seems a bit 'weird' to me, as usually the $2$-norm is bounded above by the infinity norm, not the other way as here.

I would greatly appreciate any help.

Reading some article a while ago I read the following: (here $H^2$ represents the Hardy space)

Let $f\in H^2$ be such that $f(0)=1$, and let $0<\lvert\lambda\rvert<1$, then $$\lVert f(\lambda z)\cdots f(\lambda^{n-1}z)\rVert _\infty \lVert f\rVert _2\leq \prod_{n\geq 1} \lVert f(\lambda^{n-1}z)\rVert_2$$

for every $n\geq 1$.

This seems a bit 'weird' to me, as usually the $2$-norm is bounded above by the infinity norm, not the other way as here.

I would greatly appreciate any help.