Let $E$ be a normed vector space and $f,g: X \rightarrow E-{0}$ are two continuous functions where $\|f(x)-g(x)\| < \|f(x)\|$ for any $x \in X$.

Question

I want to prove that if $E$ is a normed vector space and $f,g: X \rightarrow E-\{0\}$ are two continuous functions where $\|f(x)-g(x)\| < \|f(x)\|$ for any $x \in X$. Then $f\simeq g$.

Let $H: X \times I \rightarrow E-\{0\}$ be defined by $H(x,t) = (1-t)f(x) + tg(x) $.

$f(x)$ and $g(x)$ are continuous so $H(x,t)$ is continuous, and $H(x,0) = f(x)$, $H(x,1) = g(x)$ so $f \simeq g$.

So I did not use at all the property $\|f(x)-g(x)\| < \|f(x)\|$, but still got a homotopy between $f$ and $g$. Is my proof wrong anywhere? If it is how can I prove the proposition?

Answer 1

$H(t,x)=0$ is equivalent to $(1-t)f(x)+tg(x)=0$, this is equivalent to $f(x)=t(f(x)-g(x))$, we deduce that $\|f(x)\|=t\|f(x)-g(x)\|$ contradiction since $\|f(x)\|>\|f(x)-g(x)\|\geq t\|f(x)-g(x)\|$ since $0\leq t\leq 1$. We deduce that $H$ is well defined.

Answer 2

The thing for the homotopy to work is that $0$ is not on the segment $[f(x) g(x)]$. If it were then you would have $||f(x)-g(x)||= ||f(x)||+ ||g(x)||$ . So as long as you have the inequality $$||f(x)-g(x)||< ||f(x)||+ ||g(x)||$$ for all $x$ you are OK.