Norm of an integral operator involving linear and exponential terms Let's consider the operator $A:(C[0,1],\lVert \cdot \rVert_1) \longrightarrow (C[0,1],\lVert \cdot \rVert_1)$ defined by
$$
Ax(t)=\int_{0}^{t}{tse^sx(s)ds}
$$
I want to compute $\lVert A  \rVert_1$. First of all, $A$ is indeed a bounded operator:
$$
\lVert Ax  \rVert_1= \int_{0}^{1}{|Ax(t)|dt} \leq \int_{0}^{1}{\int_{0}^{t}tse^s |x(s)| ds dt} \leq \int_{0}^{1}{\int_{0}^{1} t \cdot 1 \cdot e^1 |x(s)| ds dt}=e \cdot \lVert x \rVert_1 \int_{0}^{1}{1 dt}= e \lVert x \rVert_1 
$$
Thus, the natural candidate for the norm is $e$. The usual trick would be trying to find a function $x(s) \in C[0,1]$ such that $\lVert x \rVert=1$ and $\lVert Ax \rVert=e$. However, in this case,it's not obvious for me what function $x(s)$ would satisfy that condition. The first natural candidate would be the constant function $x \equiv 1$. However, this function doesn't satisfy that $\lVert Ax \rVert=e$. Next, I assumed $x(s)>0$ for all $s$ and used Fubini's theorem which led me to the equation:
$$
\frac{1}{2} \int_{0}^{1}{(s-s^3)e^sx(s)ds}=e
$$
This, in the hope of obtaining a "friendlier" condition on the function $x(s)$ that would allow me to obtain what I want. As you can see, this didn't improve the situation haha :(.
Any suggestion?
In advance thank you very much.
 A: Let
$$
c=\max\Big\{\frac12\,(s-s^3)e^s:\ 0\leq s\leq1\Big\}.
$$
This number, which is around $0.36085$, is not pretty to calculate but can be calculated exactly as all it involves is finding the real roots of the degree-three polynomial $s^3+3s^2-s-1$ and then evaluating the function on said root.
We have, as you calculated,
$$
\|Ax\|_1\leq\frac12\int_0^1 (s-s^3)e^s\,|x(s)|\,ds\leq c\,\int_0^1|x(s)|\,ds=c\,\|x\|_1. 
$$
So $\|A\|\leq c$, and we can then see that $\|A\|=c$. Indeed, if $s_0$ is the point such that
$$
\frac12\,(s_0-s_0^3)e^{s_0}=c,
$$
fix $\varepsilon>0$  and let $\delta>0$ such that $\frac12\,(s-s^3)e^s\geq c-\varepsilon$ for all $s\in[s_0-\delta,s_0+\delta]$. Define
$$
x=\Big(\frac1{\delta}-\frac1{\delta^2}|x-s_0|\Big)\,1_{[s_0-\delta,s_0+\delta]}.
$$
As awful as it looks, the shape of this is simply a triangle of area $1$, centered at $s_0$ and with base $2\delta$. We have $\|x\|_1=1$, and
\begin{align}
\|Ax\|_1&=\frac 1{2}\,\int_0^1(s-s^3)e^s\,x(s)\,ds\\[0.3cm]
&=\frac 1{2}\,\int_{s_0-\delta}^{s_0+\delta}(s-s^3)e^s\,x(s)\,ds\\[0.3cm]
&\geq(c-\varepsilon)\,\int_{s_0-\delta}^{s_0+\delta} x(s)\,ds. \\[0.3cm]
&=c-\varepsilon. 
\end{align}
As this can be done for any $\varepsilon>0$, we get that $\|A\|\geq c$, and so $\|A\|=c$.
A: The operator is of the form
$$(Ax)(t)=\int\limits_0^1K(s,t)x(s)\,ds, \quad K(s,t)\ge 0 $$
We will show that $$\|A\|=\sup_{0\le s\le 1}\int\limits_0^1K(s,t)\,dt$$ First of all since $K(s,t)\ge 0$
we have $\|Ax\|_1\le \|A|x|\|_1$, therefore it suffices to restrict to nonnegative functions $x.$
Then
$$\|Ax\|_1=\int\limits_0^1\left (\int\limits_0^1K(s,t)x(s)\,ds\right )\,dt
\\ =\int\limits_0^1\left (\int\limits_0^1K(s,t)\,dt \right )x(s)\,ds
$$
Denote $\displaystyle k(s)=\int\limits_0^1K(s,t)\,dt.$ Then
$$\|Ax\|_1=\int\limits_0^1 k(s)x(s)\,ds$$
By duality between $\|\cdot\|_1$ and $\|\cdot\|_\infty $ norms we get
$$\sup_{\|x\|_1\le 1}\|Ax\|_1=\sup_{\|x\|_1\le 1}\int\limits_0^1 k(s)x(s)\,ds=\sup_{0\le s\le 1}k(s)=\sup_{0\le s\le 1}\int\limits_0^1K(s,t)\,dt$$
In the OP case we have
$$K(s,t)=tse^s\mathbf{1}_{[0,t]}(s)\,\qquad k(s)=\int\limits_0^1 tse^s\mathbf{1}_{[0,t]}(s)\,dt={1\over 2}(s-s^3)e^s$$
Remark A similar reasoning shows that for $(C[0,1],\|\cdot\|_\infty)$ we get
$$\|A\|= \sup_{0\le t\le 1}\int\limits_0^1K(s,t)\,ds$$
