Is the subset of C([0,1]) with f(1/2) = 0 a subspace?

[Alex6200]

Subspace of a Function?!?

Homework Statement

{f $$\in$$ C([0, 1]): f(1/2) = 0}

Is this subset of C([0,1]) a subspace?

Homework Equations

C[0,1] be the set of all functions that are continuous on [0, 1].

(f + g)(x) = f(x) + g(x)

(af)(x) = a*f(x)

The Attempt at a Solution

Okay, so if f is in the subspace of C than any linear combination of functions in the set should also be in the set. I understand that f is just the set of all functions for which f(1/2) = 0, but how am I supposed to answer the question formally?

Intuitively I understand that you might have two functions in the set, like:

f(x) = x - 1/2

g(x) = x² - 1/4

And I know that:

(f + g)(x) = x² + x - 3/4

(f + g)(1/2) = 1/4 + 2/4 - 3/4

(f + g)(1/2) = 0

Showing that it is a subspace if I just use those two functions. But how do I generalize my results to include any functions that could be in the set?

 

[Mark44]

Take any two functions f and g that are continuous on [0, 1]. It's given that f(1/2) = 0 = g(1/2). What about their sum? What willl (f + g)(1/2) be? What will a*f(1/2) be for any scalar a?

 

[Alex6200]

Take any two functions f and g that are continuous on [0, 1]. It's given that f(1/2) = 0 = g(1/2). What about their sum? What willl (f + g)(1/2) be? What will a*f(1/2) be for any scalar a?

I definitely understand the logic of what you're saying. If f(1/2) and g(1/2) are both 0, then (f + g)(1/2) will also be zero.

But I don't know how to say it formally.

 

[Mark44]

Let's give the subset a name:
W = {f $\in$ C([0, 1]): f(1/2) = 0}

If f and g are in W, what does that mean, and what can you say about (f + g)?
If f is in W, what can you say about (af), where a is a constant?

 

[Alex6200]

The way I did it was:

W = {(f in C[0, 1]): f(1/2) = 0}

f(x) and g(x) are in the subset W

if W is a subspace then h(1/2) = 0 for h(x) = a*f(x) + b*g(x)

h(1/2) = a*f(1/2) + b*g(1/2)
h(1/2) = a*0 + b*0
h(1/2) = 0

So W is a subspace.

--

Does that seem about right?

 

[Office_Shredder]

That's nearly textbook perfect. All you missed is

[if W is a subspace then h(1/2) = 0 for h(x) = a*f(x) + b*g(x)

This is true, except that's not what you want to prove. What you should have written is the converse: If h(1/2)=0 for h(x) = a*f(x) + b*g(x), then W is a subspace.

 

[Alex6200]

Thanks. Can I ask if I did another question correctly?

The question asks:

Show that if S and T are subspaces of a vector space V, then S $$\cap$$ T is also a subspace.

My Solution thus far

S $$\cap$$ T $$\subset$$ S and S is a subspace, so S $$\cap$$ T is also a subspace.

That almost seems too easy though, but it seems obvious that if a set is a subspace, then any part of that set must also be a subspace.

 

[Mark44]

The trouble is that not everything that belongs to a vector space (or subspace) is a subspace of that vector space (subspace). For example, R^2 with the usual rules for vector addition and scalar multiplication is a vector space, and the line x = 1 is in that space, but that line is not a subspace.

You have to show that if u and v are elements of S $$\bigcap$$ T, then u + v is in S $$\bigcap$$ T, and so is au, where a is any scalar (which is one way to show that 0 is in S $$\bigcap$$ T).

 

[Alex6200]

Okay, here's how I approached the problem.

u is in S $$\cap$$ T
v is in S $$\cap$$ T
a, b are scalars

If S $$\cap$$ T is a subspace, then au + bv is in S $$\cap$$ T.
If au + bv is in S $$\cap$$ T, then au + bv is in both S and T.

au + bv is in S because S is a subspace and au + bv is in T because T is a subspace.

There S $$\cap$$ T is a subspace.

Does that seem about right?

 

[Office_Shredder]

Okay, here's how I approached the problem.

u is in S $$\cap$$ T
v is in S $$\cap$$ T
a, b are scalars

If S $$\cap$$ T is a subspace, then au + bv is in S $$\cap$$ T.
If au + bv is in S $$\cap$$ T, then au + bv is in both S and T.

au + bv is in S because S is a subspace and au + bv is in T because T is a subspace.

There S $$\cap$$ T is a subspace.

Does that seem about right?

You got your if statements backwards. While

If S $$\cap$$ T is a subspace, then au + bv is in S $$\cap$$ T.
If au + bv is in S $$\cap$$ T, then au + bv is in both S and T.

Is true, what you did is prove au+bv is in both S and T, so what you should write is


S $$\cap$$ T is a subspace only if au + bv is in S $$\cap$$ T.
au + bv is in S $$\cap$$ T, only ifau + bv is in both S and T.

 

[Mark44]

Okay, here's how I approached the problem.

u is in S $$\cap$$ T
v is in S $$\cap$$ T
a, b are scalars

If S $$\cap$$ T is a subspace, then au + bv is in S $$\cap$$ T.
If au + bv is in S $$\cap$$ T, then au + bv is in both S and T.

au + bv is in S because S is a subspace and au + bv is in T because T is a subspace.

There S $$\cap$$ T is a subspace.

Does that seem about right?

To concur with Office_Shredder, but saying it a little differently, don't start off by saying "If <what you're trying to prove>". Instead, that should go at the end of your logical progression.

IOW,
if u in S $\cap$ T, v in S $\cap$ T, a, b are scalars, where S and T are subspaces of a vector space then <your argument here> ==> S $\cap$ T is a subspace.

Do you get what I'm saying?

 

[Alex6200]

Yeah. This is like my 2nd or 3rd proof so admittedly I'm a little rusty on how to formally write out the proofs.