Have You Tried The Simplest Substitution?

Algebra Level 4

$\left(f\left( x+y \right)\right)^2 \geq \left(f\left( x \right)\right)^2 + \left(f\left( y \right)\right)^2 \text{ for all } x, y \in\mathbb{R}.$

Let $f:\mathbb{R} \mapsto \mathbb{R}$ be a function that satisfy the condition above.

Evaluate $f \left( 2016 \right)$ .

The answer is 0.

1 solution

Félix de Montemar
Mar 25, 2016

If $x=0$ , then $f^{ 2 }\left( 0 \right) \ge f^{ 2 }\left( 0 \right) +f^{ 2 }\left( 0 \right) \Rightarrow 0\ge f^{ 2 }\left( 0 \right)$ .

If $y=-x$ , then $0\ge f^{ 2 }\left( 0 \right) =f^{ 2 }\left( x-x \right) \ge f^{ 2 }\left( x \right) +f^{ 2 }\left( -x \right)$ .

Since $f^{ 2 }\left( x \right) \ge 0$ for all $x\in\mathbb{R}$ , it implies that $f^{ 2 }\left( x \right) =0$ and $f\left( x \right) =0$ for all $x\in\mathbb{R}$

You are claiming the only function which meets the condition is f(x)=0 ...?

Let f be the identity function, then it satisfies your inequality, right?? So the answer is not unique...

Nathanael Case - 5 years, 2 months ago

No, the identity function doesn't satisfy this condition. Take $x=1,y=-1$ to see why.

A Former Brilliant Member - 5 years, 2 months ago

The notation $f^2(x)$ means $f(f(x))$ ,doesn't it?

If f is the identity then f^2 is also the identity function and so we just get x+y >= x+y which is always true...

I guess the problem meant to say $(f(x))^2$ instead of $f^2(x)$ ...?

Nathanael Case - 5 years, 2 months ago

@Nathanael Case – Its multiplication, not composition.

A Former Brilliant Member - 5 years, 2 months ago

Have You Tried The Simplest Substitution?

The answer is 0.

1 solution

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