Show that the inverse function of the identity $\text{id}\colon (C[0,1],\lVert\cdot\rVert_{\infty})\to (C[0,1],\lVert\cdot\rVert_1)$ is not continuous

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: $$\forall M\geq 0~\exists x\in [0,1]:\quad \sup_{x\in [0,1]}\lvert f(x)\rvert>M\int_0^1\lvert f(x)\rvert\, dx$$

• are the both spaces Banach spaces? – ahdahmanii Aug 20 '18 at 13:11
• $(C[0,1], \lVert\cdot\rVert_1)$ is no Banach space. – Rhjg Aug 20 '18 at 13:14

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 $$\lVert \text{id}^{-1}(f_n)\rVert_{\infty}=\lVert f_{n}\rVert_{\infty}> M\lVert f_{n}\rVert_1.$$

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 $\infty$.

• This function is not continuous! could you please explain to me? – ahdahmanii Aug 20 '18 at 13:20
• @The_lost I have the same problem. – Rhjg Aug 20 '18 at 13:21
• Oh, sorry, I overlooked that condition. It is trivial to construct a similar continuous example. I will fix it. – A. Pongrácz Aug 20 '18 at 13:27
• 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. – Rhjg Aug 20 '18 at 13:27
• @A.Pongrácz Very nice counter example +1 – ahdahmanii Aug 20 '18 at 13:36

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 $M$.