$T:\ L^2(0,1)\rightarrow L^1(0,1)$ find operator norm $$T:\ L^2(0,1)\rightarrow L^1(0,1)$$
$$Tf(t)=f(t^{3/2})$$
How do I find (and prove) the norm of this operator?
My solution is a bit intuitive and not rigorous at all:
I know, that $||T||=\sup_{||f||=1}||Tf||$. So I take a set of functions, that are in $L^2$ and normalize them all to 1. That is:
$$\{(2\alpha+1)\frac{1}{x^\alpha}\ |\ \alpha\in(0,\frac{1}{2}) \}$$
I take this set, because $\frac{1}{\sqrt{x}}$ does not lie in $L_2$ so I suppose, that
$$||T(\frac{2}{\sqrt{x}})||_1=8$$
is the norm I'm looking for. But I'm not sure how to prove this rigorously
 A: So, $\vert\vert Tf\vert\vert=\int_0^1f(t^{3/2})dt=\int_0^1f(u)(\frac{2}{3u^{1/3}})du$. To find the norm, we turn to the variational problem:
Maximise $I[f]={\int_0^1f(x)(\frac{2}{3x^{1/3}})dx}$ given that $J[f]=\int_0^1f(x)^2dx=1$
Calculus of Variations crossed with Lagrange multipliers gives us that we must solve $\delta I[f]-\lambda J[f]=0, J[f]=1$. This leads us to $\frac{2}{3x^{1/3}}-\lambda (2f)=0$, i.e. that our extremal solution will be a multiple of $x^{-1/3}$.
Routine verification gives that we must have $f(x)=\frac{1}{\sqrt3}x^{-1/3}$, which gives $\vert\vert Tf \vert\vert=\frac{2}{\sqrt3}$. Comparison with $f(x)=1$ confirms that this is a maximum, and we obtain $\vert\vert T\vert\vert=\frac{2}{\sqrt3}$
Note also that we can alternatively use Cauchy-Schwarz to jump straight to the solution, as $\vert\vert Tf \vert\vert = \langle f,\frac{2}{3u^{1/3}}\rangle\leq\vert\vert f\vert\vert_{L^2}\vert\vert\frac{2}{3u^{1/3}}\vert\vert_{L^2}$

[This solution is for if the codomain is also $L^2(0,1)$, based on an initial misreading of the question]
$\vert\vert Tf\vert\vert^2=\int_0^1f(t^{3/2})^2dt=\int_0^1f(u)^2(\frac{2}{3u^{1/3}})du=\int_0^1(f(u)\frac{\sqrt6}{3u^{1/6}})^2du$, so $T$ has the same norm as a multiplication operator. (Above, we make the substitution $t^{3/2}=u,dt=\frac{2du}{u^{1/3}}$)
Cauchy-Schwartz gives us that $\vert\vert Tf\vert\vert^2\leq\vert\vert f\vert\vert^2\vert\vert g\vert\vert^2$, where $g(x)=\frac{\sqrt6}{3x^{1/6}}$, and also gives us a direct route to finding that this bound is achieved (when $f=g$).
We note finally that $\vert\vert g\vert\vert^2=\int_0^1(\frac{2}{3x^{1/3}})dx=[x^{2/3}]_0^1=1$, so the operator norm of $T$ is $1$.
Interestingly, this proof gives us another result basically for free: If $h(x)$ is a $C^1$-bijection from $[0,1]$ to itself, then the operator $C_hf(x)=f(h(x))$ will have norm equal to $1$. I suspect that the $C^1$ constraint may not even be necessary - the mapping is basically a [permutation/automorphism] of $[0,1]$, and on finite-dimensional vector spaces, such transformations readily have norm 1. Does make the calculus cleaner to add the $C^1$ constraint anyhow.
