# Tag Info

## Hot answers tagged functional-analysis

3

Take your open cover of a ball of radius 5. Shrink it by a factor of 5. It's an open cover of the ball of radius 1. Therefore it has a finite subcover. Now reflate the subcover by a factor of 5. Et Voila! It covers the ball of radius 5.

3

If $\mathcal U_1 = \{ U \}$ is an open cover of the unit ball, then $\mathcal U_5 = \{ 5U \}$ is an open cover of the ball of radius $5$ where $5U = \{ 5x : x \in U \}$. And vice versa in the obvious way. Now the result follows: any open cover of the $5$-radius ball can be mapped to a cover of the unit ball, which has a finite sub-cover; then map back that ...

2

Given an open cover of the closed radius 5 ball, scale it down by a factor of $\frac15$. You obtain an open cover of the closed unit ball. By the compactness of that ball, you can pick a finite subcover. That finite subcover, scaled back up by a factor fo $5$, is a finite subcover of the original open cover of the radius 5 ball.

2

You should check the Lawrence Evans' book about PDEs. More precisely, you can check its fifth chapter. There, you will see that, under some really general assumptions on the open set $\Omega \subset \mathbb{R}^n$, you have that $C^{\infty}$ is dense on $W^{k,p}$. About the other question, you generally have that $C^{\infty}_c$ is not dense on the ...

2

I think the answer is no. At the very least, it is not true that the set of closed sets is the complement of the set of open sets. Instead, the closed sets are the complements of the open sets, which is different. Indeed, adding open sets also adds closed sets! The example I have in mind is $X=L^2([0,\pi])$, where $\sin(nx)$ is a sequence which is contained ...

2

Yes, you can even get $\operatorname{ran}(TS) = \ker S$, without taking the closure on the left. Let $A:\ker T\to H$ be any isomorphism; composing it with a projection $H\to \ker T$ we get a linear operator $B:H\to H$ such that $B(\ker T) = H$. Define $$S = T^\perp + TBT^\perp,\quad \text{ where } T^\perp = I-T$$ I leave it for you to check that ...

1

A different way (but starting like Jack D'Aurizio): If $a+b = \pi/2$, since $|u+v| \le \sqrt{2(u^2+v^2)}$, $|\sin(a)+\sin(b)| =|\sin(a)+\cos(a)| \le \sqrt{2(\sin^2(a)+\cos^2(a))} =\sqrt{2}$.

1

If $a+b=\frac{\pi}{2}$, then: $$\sin(a)+\sin(b) = \sin(a)+\cos(a) = \sqrt{2}\sin\left(a+\frac{\pi}{4}\right)$$ hence the range is $\displaystyle\color{red}{\left[-\sqrt{2},\sqrt{2}\right]}$.

1

There are only two such equivalence relations. In one, all three elements are equivalent. In the other, 1~2 (as required) but 3 is not equivalent to any element (other than itself).

1

If $B_5(x_0)$ is a closed ball with center $x_0$ and radius 5 in a Banach space, then $f(x):=5x+x_0$ maps the unit ball continuously onto $B_5(x_0)$. If the unit ball is compact, then so is $B_5(x_0)$ (as the image of a compact set under a continuous function).

Only top voted, non community-wiki answers of a minimum length are eligible