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I'm just stuck on this question. How can I represent the additive inverse of all continuous functions?

The additive inverse: For every $\overrightarrow{u}$ in V, there is a vector V denoted by $\overrightarrow{-u}$ such that $\overrightarrow{u}$ + ($\overrightarrow{-u}$) = $\overrightarrow{0}$.

Any help is appreciated.

This is a solution I found to a similiar problem earlier:

Describe the additive inverse of the vector space $P_3$ where $P_3$ is the set of all polynomials of degree 3 or below. Solution: $-(a_0 + a_1x + a_2x^2 + a_3x^3) = -a_0 - a_1x - a_2x^2 - a_3x^3$

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    Inverse or identity? And what do you mean by describe? Describe how, in what language or context? Remember that complete strangers on the internet don't have your homework sheet, and class notes to work with...2012-10-02
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    It really is the additive identity. I'll edit my question.2012-10-02
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    $f\equiv 0$, maybe?2012-10-02
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    Despite the edit, the title asks for the inverse; the body, the identity. Please bring them into agreement.2012-10-02
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    You've already done it --- the additive inverse of the function $u$ is described/represented by $-u$.2012-10-02
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    On the other hand, the title asks for the additive inverse of a *set* of functions, and that makes no sense. Addition is defined for pairs of functions, not for pairs of sets of functions.2012-10-02
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    That's what I was figuring. However, that is what the question asks for - word for word. For instance, it's simple to do so for $\mathbb R^4$, which was another question.2012-10-02

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I'm not sure if this is what you're after, but the additive identity is always given by $0\cdot\mathbf{v}$ for any $\mathbf{v}$ in the vector space.

Proof: $$\mathbf{v} = 1\cdot\mathbf{v}=(1+0)\mathbf{v} = 1\cdot\mathbf{v} + 0\cdot\mathbf{v} = \mathbf{v} + 0\cdot\mathbf{v}$$ Applying $(\mathbf{-v})$ to both sides then yields $\mathbf{0} = 0\cdot\mathbf{v}$

If the additive inverse is what you want instead, then a similar result will show that $(-1)\mathbf{v}$ is the additive inverse for $\mathbf{v}$

Proof: $$\mathbf{v} + (-1)\mathbf{v} = 1\cdot\mathbf{v} + (-1)\mathbf{v} = (1+(-1))\mathbf{v} = 0\cdot\mathbf{v} = \mathbf{0}$$ Again applying $(\mathbf{-v})$ yields $(\mathbf{-v}) = (-1)\mathbf{v}$

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    It is the additive inverse that I am being asked to find. I'm stuck trying to find it for a set of continuous functions. How can I describe it for the entire set?2012-10-02
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    I'm really not clear on what exactly you're asking here.... What would the inverse for a set in $\mathbb{R}^4$ be then? Give us an example to work with.2012-10-02
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    I posted an example of the solution to the vector space $P_3$ above.2012-10-02
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    I think just writing that the additive inverse of $f$ is $-f$ is sufficient.2012-10-02
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    Makes sense. Thanks for your help.2012-10-02