Sketching $$f(x)=4-x^2$$ and Proving its Concavity in D=[-2,2]

[Dostre]

Let D=[-2,2] and $$f:D\rightarrow R$$ be $$f(x)=4-x^2$$ Sketch this function.Using the definition of a concave function prove that it is concave (do not use derivative).

Attempt:
$$f(x)=4-x^2$$ is a down-facing parabola with origin at (0,4). I know that.

Then, how do I prove that f(x) is concave using the definition of a concave function? I got the inequality which should hold for f(x) to be concave:

For two distinct non-negative values of x u and v

$$f(u)=4-u^2$$ and $$f(v)=4-v^2$$

Condition for a concave function:

$$ \lambda(4-u^2)+(1-\lambda)(4-v^2)\leq4-[(\lambda u+(1-\lambda)v]^2$$

After expanding the inequality above I get:

$$(\lambda u-\lambda v)^2\leq(\sqrt{\lambda} u-\sqrt{\lambda} v)^2$$

I do not know what to do next.

 

[SammyS]

Let D=[-2,2] and $$f:D\rightarrow R$$ be $$f(x)=4-x^2$$ Sketch this function.Using the definition of a concave function prove that it is concave (do not use derivative).

Attempt:
$$f(x)=4-x^2$$ is a down-facing parabola with origin at (0,4). I know that.

Then, how do I prove that f(x) is concave using the definition of a concave function? I got the inequality which should hold for f(x) to be concave:

For two distinct non-negative values of x u and v

$$f(u)=4-u^2$$ and $$f(v)=4-v^2$$

Condition for a concave function:

$$ \lambda(4-u^2)+(1-\lambda)(4-v^2)\leq4-[(\lambda u+(1-\lambda)v]^2$$

You have to show that the above is true, not assume it's true.

Isn't it also required that 0 ≤ λ ≤ 1 ?

After expanding the inequality above I get:

$$(\lambda u-\lambda v)^2\leq(\sqrt{\lambda} u-\sqrt{\lambda} v)^2$$

I do not know what to do next.

 

[Dostre]

You have to show that the above is true, not assume it's true.

Isn't it also required that 0 ≤ λ ≤ 1 ?

Yeah it is required that 0 ≤ λ ≤ 1.

But, how do I prove that the below inequality is true?:

$$λ(4−u)^2+(1−λ)(4−v)^2≤4−[(λu+(1−λ)v]^2$$

 

[SammyS]

Yeah it is required that 0 ≤ λ ≤ 1.

But, how do I prove that the below inequality is true?:

$$λ(4−u)^2+(1−λ)(4−v)^2≤4−[(λu+(1−λ)v]^2$$

Without Loss of Generality, let u ≥ v .

What else do you know about u & v ?

 

[Dostre]

Without Loss of Generality, let u ≥ v .

What else do you know about u & v ?

That they are non-negative. I was working on this problem for a long time and I managed to turn the inequality:

$$ \lambda(4-u^2)+(1-\lambda)(4-v^2)\leq4-[(\lambda u+(1-\lambda)v]^2$$

into

$$\lambda (u-v)^2(1-\lambda)\leq0$$

Since 0<λ<1 the above inequality is true and the function f(x) is concave. I think it is correct.

 

[SammyS]

That they are non-negative. I was working on this problem for a long time and I managed to turn the inequality:

$$ \lambda(4-u^2)+(1-\lambda)(4-v^2)\leq4-[(\lambda u+(1-\lambda)v]^2$$

into

$$\lambda (u-v)^2(1-\lambda)\leq0$$

Since 0<λ<1 the above inequality is true and the function f(x) is concave. I think it is correct.

Why do you say u & v are non-negative?

 

[Dostre]

Why do you say u & v are non-negative?

Oh no they are not. My mistake. I was looking at another problem I am doing right now. It does not really matter whether they are negative or not for this inequality since the power of 2 makes their difference positive. But yeah they can be negative since they are distinct values of x and the the domain of x is [-2,2].