Finding subspace/continuous function

**steph3824** · July 13th 2011, 06:33 PM

Prove that the set U={f ∈ C([0,1]): f(1/2)=f(1)} is a subspace of C([0,1]).

I am not sure how to do this. I know what the necessary conditions are for a subspace but I can't quite figure out how to show them given a continuous function. Help please?

**dwsmith** · July 13th 2011, 06:47 PM

Originally Posted by steph3824

Prove that the set U={f ∈ C([0,1]): f(1/2)=f(1)} is a subspace of C([0,1]).

I am not sure how to do this. I know what the necessary conditions are for a subspace but I can't quite figure out how to show them given a continuous function. Help please?

You have to prove the subspace axioms.

U is nonempty since $f\in U$

Let $f,g\in U$

$(f+g)(1/2)=f(1/2)+g(1/2)=f(1)+g(1)=(f+g)(1)$

How about the last axiom now.

**steph3824** · July 13th 2011, 07:01 PM

Would it be (af)(1/2) = a·f(1/2) = a·f(1)?

**Drexel28** · July 13th 2011, 07:38 PM

Originally Posted by steph3824

Prove that the set U={f ∈ C([0,1]): f(1/2)=f(1)} is a subspace of C([0,1]).

I am not sure how to do this. I know what the necessary conditions are for a subspace but I can't quite figure out how to show them given a continuous function. Help please?

I don't suppose you know the fact that if $T:V\to W$ is a linear equation, then $\ker T$ is a subspace of $V$ . Well, I'm sure you can note that $C[0,1]\to\mathbb{R}:f\mapsto f(1)-f(\tfrac{1}{2})$ is a linear transformation and $\ker f=U$ .

**dwsmith** · July 13th 2011, 07:45 PM

Originally Posted by steph3824

Would it be (af)(1/2) = a·f(1/2) = a·f(1)?

Good.

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