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, Brahadeesh Nov 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 $\|f(x_n)\|\geq\delta$.

  • $\begingroup$ Where did you use f(0) = 0 ? $\endgroup$ – mimi Nov 28 '18 at 18:47
  • $\begingroup$ 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 $|f(x_n)|\geq\delta$. $\endgroup$ – Olivier Moschetta Nov 28 '18 at 18:51

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