First of all, there is clearly no bound if $0$ is an eigenvalue of $Lu=(pu')'+qu$, $u(0)=u(1)=0$. (This will not happen if $p(x)>0, q(x)\ge 0$ because then the operator is positive.) If $0\notin\sigma(L)$, then $u=L^{-1}f$, so an unsophisticated bound is $$ \|u\|_1\le\|u\|_2\le \|L^{-1}\| \|f\|_2 = \frac{\|f\|_2}{\textrm{dist}(0,\sigma(L))} $$ (while unsophisticated, this is also optimal however as a bound on $\|u\|_2$).

Or recall that $L^{-1}$ is an integral operator with kernel $$ G(x,y)=\begin{cases} u_0(x)u_1(y) & y\ge x \\ u_0(y)u_1(x) & x>y \end{cases} ; $$ here $u_0, u_1$ solve $(pu')'+qu=0$, satisfy the boundary condition $u_j(j)=0$ for $j=0,1$, and are normalized by the condition $W(u_0,u_1)=u_0 pu_1' - u_1pu_0'=1$. (This can be derived from the variation of constants formula, or you can just check that it works.)

So $u(x)=\int_0^1 G(x,y)f(y)\, dy$, and this also produces various bounds, for example $\|u\|_1\le 2\|u_0\|_1\|u_1\|_1\|f\|_{\infty}$.

First of all, there is clearly no bound if $0$ is an eigenvalue of $Lu=(pu')'+qu$, $u(0)=u(1)=0$. (This will not happen if $p(x)<0, q(x)\ge 0$ because then the operator is positive, but if $p,q$ have the same sign, then anything can happen.) If $0\notin\sigma(L)$, then $u=L^{-1}f$, so an unsophisticated bound is $$ \|u\|_1\le\|u\|_2\le \|L^{-1}\| \|f\|_2 = \frac{\|f\|_2}{\textrm{dist}(0,\sigma(L))} . \tag{1} $$

Or recall that $L^{-1}$ is an integral operator with kernel $$ G(x,y)=\begin{cases} u_0(x)u_1(y) & y\ge x \\ u_0(y)u_1(x) & x>y \end{cases} ; $$ here $u_0, u_1$ solve $(pu')'+qu=0$, satisfy the boundary condition $u_j(j)=0$ for $j=0,1$, and are normalized by the condition $W(u_0,u_1)=u_0 pu_1' - u_1pu_0'=1$. (This can be derived from the variation of constants formula, or you can just check that it works.)

So $u(x)=\int_0^1 G(x,y)f(y)\, dy$, and this also produces various bounds, for example $\|u\|_1\le 2\|u_0\|_1\|u_1\|_1\|f\|_{\infty}$.

While unsophisticated, (1) is also optimal as a bound on $\|u\|_2$, and it has the advantage that we can say something directly in terms of $p,q$. For example, if $p(x)\le -1$, $q(x)\ge 0$, then $\textrm{dist}(0,\sigma(L))\ge\pi^2$ because we can compare the quadratic form $\langle u,Lu\rangle =\int_0^1 (p|u'|^2+q|u|^2)$ with that of $L_0=-u''$, and the smallest (Dirichlet) eigenvalue of $L_0$ is $\pi^2$, with eigenfunction $u(x)=\sin\pi x$.

First of all, there is clearly no bound if $0$ is an eigenvalue of $Lu=(pu')'+qu$, $u(0)=u(1)=0$. If $0\notin\sigma(L)$, then $u=L^{-1}f$, so an unsophisticated bound is $$ \|u\|_1\le\|u\|_2\le \|L^{-1}\| \|f\|_2 = \frac{\|f\|_2}{\textrm{dist}(0,\sigma(L))} . $$

Or recall that $L^{-1}$ is an integral operator with kernel $$ G(x,y)=\begin{cases} u_0(x)u_1(y) & y\ge x \\ u_0(y)u_1(x) & x>y \end{cases} ; $$ here $u_0, u_1$ solve $(pu')'+qu=0$, satisfy the boundary condition $u_j(j)=0$ for $j=0,1$, and are normalized by the condition $W(u_0,u_1)=u_0 pu_1' - u_1pu_0'=1$. (This can be derived from the variation of constants formula, or you can just check that it works.)

So $u(x)=\int_0^1 G(x,y)f(y)\, dy$, and this also produces various bounds, for example $\|u\|_1\le 2\|u_0\|_1\|u_1\|_1\|f\|_{\infty}$.

First of all, there is clearly no bound if $0$ is an eigenvalue of $Lu=(pu')'+qu$, $u(0)=u(1)=0$. (This will not happen if $p(x)>0, q(x)\ge 0$ because then the operator is positive.) If $0\notin\sigma(L)$, then $u=L^{-1}f$, so an unsophisticated bound is $$ \|u\|_1\le\|u\|_2\le \|L^{-1}\| \|f\|_2 = \frac{\|f\|_2}{\textrm{dist}(0,\sigma(L))} $$ (while unsophisticated, this is also optimal however as a bound on $\|u\|_2$).

Or recall that $L^{-1}$ is an integral operator with kernel $$ G(x,y)=\begin{cases} u_0(x)u_1(y) & y\ge x \\ u_0(y)u_1(x) & x>y \end{cases} ; $$ here $u_0, u_1$ solve $(pu')'+qu=0$, satisfy the boundary condition $u_j(j)=0$ for $j=0,1$, and are normalized by the condition $W(u_0,u_1)=u_0 pu_1' - u_1pu_0'=1$. (This can be derived from the variation of constants formula, or you can just check that it works.)

So $u(x)=\int_0^1 G(x,y)f(y)\, dy$, and this also produces various bounds, for example $\|u\|_1\le 2\|u_0\|_1\|u_1\|_1\|f\|_{\infty}$.