# What does “exercise” mean here? [closed]

REPORTING on big international summits is often an exercise in drudgery.what does exercise mean here?

REPORTING on big international summits is often an exercise in drudgery.what does exercise mean here?

So, the sentence is:

We are an exercise in cognitive dissonance.My question is what does it mean to be "an exercise" in this sentence?

Should I have texted him

Do you have an exercise tonight?using present tense instead of present continuous?

When dividends exceed the time value left, then it is profitable to early exercise a call.So, when to early exercise a put?

I could not find "exercise government" in dictionaries.Could it be that "exercise government power" was the original intent?

All of the sentences below convey the meaning of compulsion of exercise to be carried out in three months.a) This exercise has to be carried out in three months.

Could I then write:

He went into exercise.However, it might be acceptable in a context where exercise meant a long-term regimen: He went into exercise to improve his poor health.

Would exercise be one of those activities?Similarly, he is permitted to read the newspaper a bit

and do a little exercise before prayer.

Does moderate exercise correlate with learning ability and memory retention?From Exercise and the brain: something to chew on:

Evidence is accumulating that exercise has profound benefits for brain

function.

Chapter $3$ of Functional Analysis, Sobolev Spaces and Partial Differential Equations by Haim Brezis constructs and explains the Weak and Weak$^*$ topologies over a Banach Space $E$.The most remarkable result of these topologies is the Banach-Alaoglu-Bourbaki, which asserts that for the weak$^*$ topology (Over $E^*$), the unit ball: $$B_{E^*}=\{f \in E^* | \ ||f|| \leq 1 \}$$ is compact.

In their 1983 JFA paper Brezis and Lieb have shown, among many other things, a Poincaré-type inequality: in the case of a harmonic function $f$ on a bounded domain $\Omega$, their inequality ((3.14) in the paper) states that the $L^2(\Omega)$-norm of $f$ can be estimated by the $L^2(\partial \Omega)$-norm of its trace on $\partial \Omega$ (times a constant only depending on $\Omega$).

Dear Mathoverflow'ers,

I am interested in the following equation:

$-\Delta u = u^{p-1} + \lambda u$ in $ \Omega$ with $ u=0 $ on $ \partial \Omega$.I realize that this is somewhat of an ill formed question and may not be suitable for mathoverflow.

I have a question about a step in proving $(M^\perp)^\perp=\overline{M}$ where $M$ is a linear subspace of a normed vector space $E$.And $M^\perp=\{f\in E^*|\langle f,x\rangle =0\}$

This is the proof from Brezis book:

Proof: Assuming the other direction, show $\subset$ direction.

As far as I know, there are two different definitions for pseudomonotone function.Brezis in 1968:

Pseudomonotone function in the sense of Brezis:

Let $V$ be a reflexive Banach space with its dual space $V^\ast$.

I am studying compact embedding in Brezis' books and I faced the following problem:

I have done the problem with $p > N$ and $p < N$, but i don't really know why the case $p = N$ reduces to the case $p < N$?Thanks for your ideas.

I'm studyin Brezis' PDE with a property:

What about $p = \infty$?I think that when $p = \infty$, the statement is wrong but I can't give an example.

Consider $H$ to be Hilbert space and $V$ to be a Banach space.Brezis says: there is a canonical map $T\colon H^*\to V^*$ that is the restriction to $V$ of continuous linear functionals $\varphi$ on $H$, i.

Determine

$$N^\bot=\{x\in E : \langle f, x\rangle = 0 \quad \forall f\in N \}$$

and

$$N^{\bot\bot}=\{f\in E^* : \langle f, x\rangle = 0 \quad \forall x\in N^\bot \}.$$

Check that $N^{\bot\bot} \ne N^\bot$?

I made up an exercise to my self for improving my normalization skills.I will describe the exercise and my solution.