# If $T(x) = 0$ implies that $S(x) = 0,$ show that there is a constant $c$ such that $S(x) = c \cdot T(x)$ for all $x \in X.$

Let $X$ be a normed vector space and suppose $T$ and $S$ are in $X^*.$ If $T(x) = 0$ implies that $S(x) = 0,$ show that there is a constant $c$ such that $S(x) = c \cdot T(x)$ for all $x \in X.$

I somehow relate it to the equivalent of two norms, as the definition gives us a constant. However, I do not know how to solve this quetsion.

Any hint would be appreciated.

• Hint: Suppose that there exists $u$ such that $T(u)\not =0$. For $x\in X$, put $y=x-\frac{uT(x)}{T(u)}$. Commented Jun 5, 2017 at 7:23
• What is the motivation of getting $y = x - \frac{uT(x)}{T(u)}$? Commented Jun 5, 2017 at 7:44
• A good exercise for you is that, let $T:X \rightarrow R^n$ and $S:X \rightarrow R^m$, with same property you said , try guess and formulate the conclusion and then try to prove it ... Commented Jun 5, 2017 at 8:05

I constructed the following answer based on Kelenner's comment.

If $T(x) = 0$ for all $x \in X,$ then let $c = 0.$ This gives us that $S(x) = 0 = c \cdot T(x).$

Suppose that there exists $u \in X$ such that $T(u) \neq 0.$ Let $y = x - \frac{u \cdot T(x)}{T(u)}.$ Note that $T(y) = 0.$ By assumption, we have $S(y) = 0.$ Note that $S(y) = S(x) - \frac{S(u) \cdot T(x)}{T(u)} = 0.$ This implies that $S(x) = \frac{S(u)}{T(u)} \cdot T(x).$

• Well done ! (+1) Commented Jun 5, 2017 at 8:18