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Consider $C[0,1]$, the set of all continuous functions $f\colon [0,1]\to\mathbb{R}$, as well as the norms $$ \lVert f\rVert_{\infty}:=\sup_{x\in [0,1]}\lvert f(x)\rvert, $$ $$ \lVert f\rVert_1:=\int_0^1\lvert f(x)\rvert\, dx. $$

I am a bit lost to show that for the identity map given in the title (which is bijective, continuous and linear), its inverse function is not continuos.


I think one way to show this is to find an example $f\in C[0,1]$ with the property: $$ \forall M\geq 0~\exists x\in [0,1]:\quad \sup_{x\in [0,1]}\lvert f(x)\rvert>M\int_0^1\lvert f(x)\rvert\, dx $$

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  • $\begingroup$ are the both spaces Banach spaces? $\endgroup$
    – ahdahmani
    Aug 20, 2018 at 13:11
  • $\begingroup$ $(C[0,1], \lVert\cdot\rVert_1)$ is no Banach space. $\endgroup$
    – Rhjg
    Aug 20, 2018 at 13:14

3 Answers 3

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Why not considering $f_n(x):=x^n$?

Then $\lVert f_n\rVert_1=\frac{1}{n+1}\to 0$ as $n\to\infty$, while $\lVert f_n\rVert_{\infty}=1$ for all $n\geq 1$.

Hence, $\forall M\geqslant 0~\exists f_n\in C[0,1]$ with $n$ large enough such that $$ \lVert \text{id}^{-1}(f_n)\rVert_{\infty}=\lVert f_{n}\rVert_{\infty}> M\lVert f_{n}\rVert_1. $$

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Exactly, your suggestion is the right proof strategy. So you show that the inverse of the unit ball is not contained in any ball.

Let $f_c(x)$ be the function for all $0<c<1$ such that $f_c(x)= \begin{cases} \frac{2}{c}-\frac{2x}{c^2} \,\, \text{if} \,\, x\in [0,c] \\ 0 \,\, \text{otherwise} \end{cases}$.

These are in the unit ball w.r.t the $L^1$ norm, but as $c\rightarrow 0$, their supremum norm tends to $\infty$.

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    $\begingroup$ This function is not continuous! could you please explain to me? $\endgroup$
    – ahdahmani
    Aug 20, 2018 at 13:20
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    $\begingroup$ @The_lost I have the same problem. $\endgroup$
    – Rhjg
    Aug 20, 2018 at 13:21
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    $\begingroup$ Oh, sorry, I overlooked that condition. It is trivial to construct a similar continuous example. I will fix it. $\endgroup$ Aug 20, 2018 at 13:27
  • $\begingroup$ Whats with $f_n(x):=x^n$ for $n\geq 1$? Then for large $n$, the integral is bounded by $1$ but the supremum is unbounded. $\endgroup$
    – Rhjg
    Aug 20, 2018 at 13:27
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    $\begingroup$ @A.Pongrácz Very nice counter example +1 $\endgroup$
    – ahdahmani
    Aug 20, 2018 at 13:36
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Recall that a linear map is continuous if and only if it is bounded. We're looking at the identity map from $(C[0, 1], \| \cdot \|_1)$ to $(C([0, 1], \| \cdot \|_\infty)$, meaning that we need to show that no bound $M$ exists for this map. That is, for all $M$, there exists some $f$ in the unit ball of $(C[0, 1], \| \cdot \|_1)$ (i.e. $\|f\|_1 \le 1)$ such that $\|f\|_\infty > M$.

Essentially, you need a continuous function that has very large maxima, but whose absolute integral is still $1$. I'd suggest some kind of piecewise linear bump functions. Connect the points $(0, 0)$, $(1/M, M)$, $(2/M, 0)$, $(1, 0)$ into a piecewise linear function. Then, the function is positive, and the integral under it is $1$, but the supremum of the function is $M$.

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