Is |f(1)| = \inf\{\sup\{|f(t)+m(t)|: t \in [0,1]\} : m \in M\}?

[.....]

Homework Statement

f is a continuous function on [0,1]

M is the set of continuous functions on [0,1] which are 0 at 1 ... i.e. for all m in M, m(1) = 0

I want to know if it's true that

$$|f(1)| = \inf\{\sup\{|f(t)+m(t)|: t \in [0,1]\} : m \in M\}$$

The Attempt at a Solution

So... you're choosing t to make it as big as possible and choosing m to make it as small as possible...

If f(1)=0 then you could choose m = -f and then for any t f+m would be zero, so the whole thing would be zero and the equation would be true...

But I'm not sure about the f(1) not equal to 0 case...

 

[rasmhop]

Try inserting:
$$m(t) = tf(1)-f(t)$$
The idea is basically the same as yours except that instead of keeping |f(t)+m(t)| constantly at 0 we make it increasing so it has supremum at t=1.