prove that if t is bounded, everywheredefined, an isometry, then ran t i

要证明如果 t 是一个有界的、处处有定义的等距映射，那么 ran t 是全空间 V。首先，我们需要了解等距映射的性质。

一个映射 t 称为等距映射，如果它保持了两个向量之间的距离。也就是说，对于 V 中的任意两个向量 u 和 v，有 ||t(u) - t(v)|| = ||u - v||，其中 ||.|| 是向量的范数。

假设 t 是有界的，这意味着对于 V 中的任意向量 u，有 ||t(u)||≤ M，对于某个 M>0。从而对于任意 v∈ V，我们有 ||t(u) - t(u+v)|| = ||u - (u+v)|| = ||-v|| = ||v||，因此 ||t(u+v)|| = ||t(u) - t(u+v)||≤ ||u+v|| ≤ ||u|| + ||v||。

通过以上的不等式，我们可以证明 t 是一个开映射。如果 t 不是开映射，那么存在一个向量 u0∈ V 和一个正实数 ε>0，使得对于所有的向量 v，满足 ||v|| < ε，有 ||t(u0 + v)|| ≥ ε。但是，根据上面的不等式，当取 v = -u0 时，我们有 ||t(u0)|| = ||t(0)|| = ||t(u0 - u0)|| = ||u0 - u0|| = 0，这与 t 是一个等距映射矛盾。

由于 t 是一个开映射，那么它必定将整个 V 映射到它的像空间上。因此，ran t = t(V) = V，即 ran t 是全空间 V。

综上所述，如果 t 是一个有界的、处处有定义的等距映射，则 ran t 是全空间 V。

相关问题

Prove that if a1, a2,..., an are n ≥ 2 integers such that ai ≡ 1 (mod 3) for every integer i (1 ≤ i ≤ n), then a1a2 ··· an ≡ 1 (mod 3)

To prove this statement, we can use mathematical induction.

For n = 2, we have a1a2 ≡ 1 (mod 3) since both a1 and a2 are congruent to 1 modulo 3.

Assume the statement is true for n = k. That is, if a1, a2, ..., ak are integers such that ai ≡ 1 (mod 3) for every integer i (1 ≤ i ≤ k), then a1a2···ak ≡ 1 (mod 3).

Now, consider n = k + 1. We have ak+1 ≡ 1 (mod 3), since ak+1 is one of the given integers, and a1, a2, ..., ak are also congruent to 1 modulo 3. Thus,

a1a2···akak+1 ≡ (a1a2···ak)ak+1 ≡ 1×1 ≡ 1 (mod 3)

by the induction hypothesis.

Therefore, the statement is true for all n ≥ 2.

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dy/dx=(y-1)(y-2) Prove that if y = f(x), x ∈ R, is a solution of (1), then y = f(x + c) is also a solution, where c ∈ R is constant.

To prove that y = f(x + c) is also a solution of the differential equation dy/dx=(y-1)(y-2), we need to show that:

(d/dx)f(x+c) = (f(x+c)-1)(f(x+c)-2)

To do this, we can use the chain rule of differentiation:

(d/dx)f(x+c) = df/d(x+c) * d(x+c)/dx

Since d(x+c)/dx = 1, we can simplify this to:

(d/dx)f(x+c) = df/d(x+c)

Now, let's substitute y = f(x) into the differential equation:

dy/dx = (y-1)(y-2)

Substituting x + c for x and f(x + c) for y, we get:

df/d(x+c) = (f(x+c)-1)(f(x+c)-2)

This is the same as what we need to prove, so we have shown that y = f(x + c) is also a solution of the differential equation.