A Multiplication operator in a Hilbert space: $M_h$ is bounded and $||M_h|| \leq || h||_{\infty}$

Question

I'm trying to understand the example below, taken from Axler's Measure Integration and Real Analysis book.

How does one prove that $M_h$ is bounded and that $||M_h|| \leq || h||_{\infty}$?

I was thinking of using Holder's Inequality, $$||fh||_{1} \leq ||f||_1 || h||_{\infty}$$ However, since we're in a Hilbert space, $L^2$ has the norm given by the inner product, which is equivalent to the $||\cdot||_2$ and not $||\cdot||_1$. Also, to use something like $L^2 \subset L^1$, we need to have $\mu (X) < \infty$, which we don't in the example. Therefore, I think I'm supposed to use another inequality, but I don't see which one.

@Mittens, I don't think so, since from the example I already know that $h \in L^{\infty}$. — An old man in the sea., Jun 12 '24 at 22:12
@Mittens the condition for closure as a duplication is "question been asked AND already has an answer". I see that the answer to the question in the link proves that $M_h$ is bounded even though the question is very different. Furthermore, it doesn't seem to prove the inequality I'm looking for. — An old man in the sea., Jun 12 '24 at 22:21
This problem has been discussed before in MSE. do some search. Here is an other instance. Notice that if $h\in L_\infty(\mu)$ and $f\in L_p(\mu)$, then $\int|hf|^p,d\mu\leq|h|\infty\int|f|^p,d\mu=|h|\infty|f|^p_p$. — Mittens, Jun 12 '24 at 23:14
@Mittens thanks for the link. However, risking your wrath, the OP there clearly states that he has already proved the inequality I want, and he's looking for the inequality in the reverse sense. So, all of the work is for the reverse inequality. — An old man in the sea., Jun 12 '24 at 23:24
Also, I'm studying this by myself, so things that may seem simple/basic to you, may be very strange/odd to me. I hope you can understand. Thanks. — An old man in the sea., Jun 12 '24 at 23:26
I get it now from your inequalities in the last comment. Thanks ;) — An old man in the sea., Jun 12 '24 at 23:30

score 2 · Accepted Answer · answered Jun 12 '24 at 21:35

2

This inequality $$\int_X|fh|^2\,d\mu\leq \|h\|_\infty^2\int_X|f|^2d\mu$$ shows that $M_h$ is bounded and that $\|M_h\|\leq \|h\|_\infty$.

answered Jun 12 '24 at 21:35

uniquesolution

18,982

Which inequality are you using, Cauchy-Schwartz? – An old man in the sea. Jun 12 '24 at 22:06
If so, how come you have $||h||_{\infty}$ instead of $||h||$? – An old man in the sea. Jun 12 '24 at 22:17
@Anoldmaninthesea.: Note that the inequality comes from the assumption, that $h \in L^\infty$. So $|fh|^2 = |f|^2 |h|^2 \leq |h|^2_\infty |f|^2$. – El Dorado Jun 12 '24 at 22:30
@ElDorado but what allows one to do $||h||<||h||{\infty}$, where $||\cdot||{\infty}$ is the essential supremum? Looking at this question answers, we need to assume that $h \in L^2$ or something similar to $\mu(X)\leq \infty$... – An old man in the sea. Jun 12 '24 at 23:19
@ElDorado https://math.stackexchange.com/questions/242779/limit-of-lp-norm?noredirect=1&lq=1 – An old man in the sea. Jun 12 '24 at 23:19

A Multiplication operator in a Hilbert space: $M_h$ is bounded and $||M_h|| \leq || h||_{\infty}$

1 Answers1