Vtyjkyuk

Consider $C[0,1]$, the set of all continuous functions $fcolon [0,1]tomathbbR$, as well as the norms
$$
lVert frVert_infty:=sup_xin [0,1]lvert f(x)rvert,
$$
$$
lVert frVert_1:=int_0^1lvert f(x)rvert, dx.
$$

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 $fin C[0,1]$ with the property:
$$
forall Mgeq 0~exists xin [0,1]:quad sup_xin [0,1]lvert f(x)rvert>Mint_0^1lvert f(x)rvert, dx
$$

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 $f_c(x)= begincases frac2c-frac2xc^2 ,, textif ,, xin [0,c] \ 0 ,, textotherwise endcases$.

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

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

Then $lVert f_nrVert_1=frac1n+1to 0$ as $ntoinfty$, while $lVert f_nrVert_infty=1$ for all $ngeq 1$.

Hence, $forall Mgeqslant 0~exists f_nin C[0,1]$ with $n$ large enough such that
$$
lVert textid^-1(f_n)rVert_infty=lVert f_nrVert_infty> MlVert f_nrVert_1.
$$

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 $M$.