Proving that a given function $f^*$ is the best least square approximation In De Boor (1972) it is stated that

Let be $\$ $  a finite dimensional linear space of functions defined on the interval $[a,b]$. We are searching for the best approximation from $\$$ to $g$.
The function $f^*$ is a best approximation from $\$$ to $g$ with respect to the $\mathcal{L}_2$ norm if and only if the function $f^*$ is in $\$$ and the error term $g-f^*$ is orthogonal to $\$$.

To show that the double implication holds the author provides the following statement (I add my own procedure because it's not given in the book).
For any function $f\in \$\ $ we have that
$$\|g-f\|_2^2=\|g-f^*+f^*-f\|_2^2=\|g-f^*\|_2^2+2\langle f^*-f,g-f^*\rangle+\|f^*-f\|_2^2$$
If condition  is satisfied we have
$$\|g-f\|_2^2=\|g-f^*\|+\|f^*-f\|_2^2\geq\|g-f^*\|_2^2$$
Which proves that $f^*$ is the best least squares approximation in $\$$
If $\langle f,g-f^*\rangle \neq 0$
we have that by letting $tf:=f^*-f$
$$\|g-f\|_2^2=\|g-(f^*+tf)+(f^*+tf)-f\|_2^2$$
$$=\|g-(f^*+tf)\|_2^2+2\langle 2tf,g-f^*-tf\rangle+\|2tf\|_2^2$$
$$=\|g-(f^*+tf)\|_2^2+4\langle tf,g-f^*\rangle$$
Given that for all nonzero $t$ of the same sign as $\langle f,g-f^*\rangle$ and sufficiently close to $0$.
$$2t\langle f,g-f^*\rangle>\|tf\|_2^2$$
This completes the proof since it implies that
\begin{equation}
\|g-(f^*+tf)\|_2^2<\|g-f^*\|_2^2
\end{equation}
and hence $f^*$ is not the best approximation. 
I am not sure to understand the reason why it holds that $$2t\langle f,g-f^*\rangle>\|tf\|_2^2$$
Is it because $t^2$ goes to zero faster than $t$.
Is there some way to show it more formally?
Thanks in advance.
 A: If $t$ has  same sign as $\langle f, g-f^{*}\rangle$ the inequality you want is $2|t||\langle f, g-f^{*}\rangle| >|t|^{2} \|f\|_2^{2}$. This is true whenever $|t| <\frac {2|\langle f, g-f^{*}\rangle|} { \|f\|_2^{2}}$.
A: I would like to suggest a slightly different proof, because I don't feel well with the argument " by letting  $tf:=f^*-f$".
Assume $\langle f,g-f^*\rangle\neq 0$ for some $f\in U$ (I call the subspace $U$), then for all $t\in\mathbb{R}$ one has:
$$||g-f^*||_2^2=||g-f^*+tf-tf||_2^2=||g-(f^*+tf)||_2^2+2\langle g-(f^*+tf),tf\rangle +||tf||_2^2$$
$$=||g-(f^*+tf)||_2^2+2t\langle g-f^*,f\rangle -2t^2\langle f,f\rangle+||tf||_2 ^2$$
$$=|||g-(f^*+tf)||_2^2+2t\langle g-f^*,f\rangle-t^2||f||_2^2.$$
And now choose $t$ so that 
$$2t\langle g-f^*,f\rangle-t^2||f||_2^2>0$$
which in case that $t$ has the same sign as $\langle g-f^*,f\rangle$ is equivalent to
$$|t|<\frac{2|\langle g-f^*,f\rangle|}{||f||_2^2}.$$
(Note that $\langle g-f^*,f\rangle\neq 0$ is essential to that.)
Then:
$$|||g-(f^*+tf)||_2^2<||g-f^*||_2^2$$
which contradicts the best-approximation-property.
