How to show this norm is less than 1?

Question

If $f$ and $g$ are linearly independent vectors in Hilbert space H with $||f||=1=||g||,$ then how to show that $||tf+(1-t)g||<1, \forall \hspace{0.1 cm} 0<t<1$

Answer 1

The triangle inequality tells us that

$$\|tf+(1-t)g\|\leq t\|f\|+(1-t)\|g\|=t+1-t$$

Now the equality holds if and only if $tf$ and $(1-t)g$ are linearly dependant This means $\exists k$ such that $tf=k(1-t)g$ but this means $f$ and $g$ are linearly dependant and so for $f$ and $g$ linearly independent one has

$$\|tf+(1-t)g\|\lt t\|f\|+(1-t)\|g\|=t+1-t$$