# Convergence in a normed vector space - Linear operator [closed]

Having $$X$$ a normed vector space. If $$f$$ is a linear operator from $$X$$ to $$ℝ$$ and is not continuous in $$0$$ (element of $$X$$) , how can we show that there exists a sequence $$x_n$$ that converges to $$0$$ for which we have $$f(x_n) = 1$$ (for all $$n$$ element of $$ℕ$$).

Any help would be greatly appreciated, thank you.

## closed as off-topic by John B, Cesareo, Chinnapparaj R, user10354138, BrahadeeshNov 26 '18 at 8:22

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By definition of a limit, construct a sequence $$x_n$$ such that $$x_n\rightarrow 0$$ but $$\|f(x_n)\|\geq\delta>0$$. Here we use $$f(0)=0$$. Then rescale by letting $$u_n=\dfrac{x_n}{\|f(x_n)\|}$$ Clearly $$\|f(u_n)\|=1$$ by linearity of $$f$$ but $$u_n\rightarrow 0$$ since $$x_n$$ does and $$\|f(x_n)\|\geq\delta$$.
• Since $f$ is not continuous at $0$ then there exists $\delta>0$ such that for every $n\in\mathbb{N}$ there exists $x=x_n$ with $|x_n|\leq 1/n$ and $|f(x_n)-f(0)|\geq\delta$. Since $f(0)=0$ this gives $|f(x_n)|\geq\delta$. – Olivier Moschetta Nov 28 '18 at 18:51